-1 converted to an unsigned integer type is guaranteed to produce the largest possible value for that unsigned type. Otherwise, if the new type is unsigned, the value is converted by 2) A processor that uses other binary representations of numbers (1's complement? Does the compiler make optimization when the result of expression fits in integer data type? Where does the idea of selling dragon parts come from? Well, we could live with that. "implementation detail" would be "implementation defined" in C++ parlance. Luckily, C2x will solve the `uint8_t` and `uint16_t` problem with `_BitInt(8)` and `_BitInt(16)`, which dont promote to `int` automatically. A simple function to convert from signed to unsigned is the key: template<typename T> auto to_unsigned(T value) { return std::make_unsigned_t<T> (value); } This function alone could raise UB if we apply to a negative value, but all the positive signed values fit in the corresponding unsigned type, and the conversion maintain the value intact. But apparently this is not limited to the value of said integer, it can also dramatically impact the code flow. representable values for its type, the behavior is undefined. Why are these constructs using pre and post-increment undefined behavior? We do not currently allow content pasted from ChatGPT on Stack Overflow; read our policy here. Why does the USA not have a constitutional court? Do non-Segwit nodes reject Segwit transactions with invalid signature? Unsigned integer arithmetic does not overflow because paragraph 6.2.5/9 applies, causing any unsigned result that otherwise would be out of range to be reduced to an in-range value. rev2022.12.11.43106. 2 The usual arithmetic conversions are performed on operands of arithmetic or enumeration type, 8.10 Equality operators [expr.eq] The integral promotion results in non-portable code. The surprising result occurs due to integral promotion. Can anyone explain this code behaviour in c++? A computation involving unsigned operands can never overflow, because a result that cannot be represented by the resulting unsigned integer type is reduced modulo the number that is one greater than the largest value that can be represented by the resulting type. What properties should my fictional HEAT rounds have to punch through heavy armor and ERA? (x | y) - y why can't it simply be x or even `x | 0`, Store an int in a char buffer in C and then retrieve the same, Runtime error in a program supposed to convert a float to a byte array, Bypassing an unsigned addition overflow detected by CBMC. I think they could have chosen a clearer wording though. When evaluating the conditional, the left hand side (sum) is promoted to type int, and the right hand side (the summation 65536) is already type int. Dual EU/US Citizen entered EU on US Passport. There is a lot of code out there that relies on this behaviour. here's the link: This makes unsigned integer types a special case. Its somewhatcontroversial whether compilers really oughtto ever do this, but the reality is that in the present day its an extremely common optimization technique, and nothing in the C/C++ standards forbids it. 6 If both operands are of arithmetic or enumeration type, the usual arithmetic conversions are performed on both operands. Browse other questions tagged, Where developers & technologists share private knowledge with coworkers, Reach developers & technologists worldwide. One approach might be using the top bit as padding and zeroing it after each operation. Unsigned integers, declared unsigned, shall obey the laws of arithmetic modulo 2^n where n is the number of bits in the value representation of that particular size of integer. The effect is that unsigned types smaller than unsigned int will be (manually) promoted to unsigned int, and unsigned types larger than unsigned int will be unchanged. reduced modulo the number that is one greater than the largest value that can be Is this a clang optimizer bug or an undefined behavior in C? Can virent/viret mean "green" in an adjectival sense? How do I detect unsigned integer overflow? Also, as suggested by nategoose, we could just declare the sum as, One could achieve such optimizations without requiring integer overflow to be Undefined Behavior, if one specifies that the result of an operation yielding a value outside the range of, where values may be held in registers longer than, This one is really old. The wrap-around is just part of the normal unsigned integer behavior and not seen as overflow: (2^n here of course means 2 raised to the power of n) undefined behavior - there are no restrictions on the behavior of the program. @R: 2's complement isn't the only issue though - e.g. One could argue that unsigned integers should behave consistently, because why not assume that a + b >= a and a + b >= b for unsigned integers in the optimizer (similar to concluding a + b > 0 for a > 0 and b > 0 in J-16 SDiZ's example for signed integers)? I find all this fuss that we have gone through (and still go) with 32 vs. 64 bit portability a real shame for the whole profession. Whenever you use a variable with unsigned type smaller than unsigned int, add 0u to it within parentheses. type is reduced modulo to the number that is one greater than the Well only signed overflow is undefined behavior. When a value with integer In my eyes this is more of an oddity in the C spec. Save. By contrast, Signed numbers are most often represented using two's complement but other choices are possible as described in the standard (section 6.2.6.2). how does c compiler handle unsigned and signed integer? Casts between signed and unsigned integer types of the same width are free, if the CPU is using 2's compliment (nearly all do). represented by the resulting unsigned integer type is reduced modulo Where developers & technologists share private knowledge with coworkers, Reach developers & technologists worldwide, The choice of words in the standard is unfortunate. The following code might be quite surprising: Heres a link to it on wandbox if you want to try it out. [] The fact that a two's complement representation is used for those signed types does not mean that arithmetic modulo 2^n is used when evaluating expressions of those types. They wont be protected by any well-defined behavior of the original unsigned type, since after promotion the types are no longer unsigned. They didn't agree on what signed overflow should do, so that did not get in the standard. The undefined behavior bits in the specification involve some compiler optimization. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Putting aside the success, C++ has all the quirks and flaws and many many more. Function Description _Exit() Exit the currently-running program and don't look back abort() Abruptly end program execution abs() Compute the absolute value of an integer aligned . This may incur minor performance losses when making comparisons (for instance, now. With unsigned numbers of type unsigned int or larger, in the absence of type conversions, a-b is defined as yielding the unsigned number which, when added to b, will yield a. You can use the following helper class to get a safe type that you can use as a safe destination type for explicit casts on your unsigned (or generic) integer types during mathematical operations and comparisons. So, why should someone avoid causing it? Both references are correct, but they do not address the same issue. It means that the implementation is allowed to do whatever it likes in that situation. , unsigned int 32 , 31 - undefined. Why the assembly code for unsigned and signed arithmetic operation are the same? . If unsigned values were merely storage-location types and not intermediate-expression types (e.g. The problem is that signed integer overflow causes undefined behaviour. "casts from unsigned -> signed int are well defined": This isn't correct; converting from unsigned to signed yields an. Certainly depends if you write your code for yourself or if you expect it to end up in a library or so. How many transistors at minimum do you need to build a general-purpose computer? We end up with the unexpected output sum = one + max, but sum != one + max". By clicking Post Your Answer, you agree to our terms of service, privacy policy and cookie policy. What is this fallacy: Perfection is impossible, therefore imperfection should be overlooked. The standard does effectively guarantee that typesint, unsigned int, long, unsigned long, long long,andunsigned long long will never be promoted. Where in the C99 standard does it say that signed integer overflow is undefined behavior? Is unsigned integer subtraction defined behavior? It means that you can't alter the sign of a unsigned calculation, but it can still produce unexpected results. So it would be implementation-defined instead. I can't really think of any situation where the overflow behaviour is actually useful @sleske: Using decimal for human-readability, if an energy meter reads 0003 and the previous reading was 9995, does that mean that -9992 units of energy were used, or that 0008 units of energy were used? A non-exhaustive list of types that mightbe promoted is, char, unsigned char, signed char, short, unsigned short, int8_t, uint8_t, int16_t, uint16_t, int32_t, uint32_t, int64_t, uint64_t, int_fast8_t, uint_fast8_t, int_least8_t, uint_least8_t,int_fast16_t, uint_fast16_t, int_least16_t, uint_least16_t,int_fast32_t, uint_fast32_t, int_least32_t, uint_least32_t,int_fast64_t, uint_fast64_t, int_least64_t, uint_least64_t. For example, for the energy meter you probably want to use a type such that overflow never occurs. Asking for help, clarification, or responding to other answers. At what point in the prequels is it revealed that Palpatine is Darth Sidious? When you work with unsigned types, modular arithmetic (also known as "wrap around" behavior) is taking place. Lets work with concrete numbers and assume your compiler uses a 16 bit unsigned short type and a 32 bit int type this is very common, though not universal. Central limit theorem replacing radical n with n. Is it correct to say "The glue on the back of the sticker is dying down so I can not stick the sticker to the wall"? If INT_MAX equals 65535, The misbehavior can even precede the overflow. However, if you are having overflows in the calculations, it is important to understand what that actually results in, and that the compiler MAY do something other than what you expect (and that this may very depending on compiler version, optimisation settings, etc). Perhaps another reason for why unsigned arithmetic is defined is because unsigned numbers form integers modulo 2^n, where n is the width of the unsigned number. So, no, I for myself I try to stick to the standards. Undefined, unspecified and implementation-defined behavior. CSAPP lablab1CSAPP1.~ &^ int bitXor(int x, int y) { return ~((~x)&(~y))&(~(x&y)); } xy0 . rev2022.12.11.43106. How do I arrange multiple quotations (each with multiple lines) vertically (with a line through the center) so that they're side-by-side? Second, the representation of a signed integer (by definition) includes representation of a sign e.g. casts from unsigned -> signed int are well defined. use integer type which can hold a result of such magnitude, for example: unsigned long long int We and our partners store and/or access information on a device, such as cookies and process personal data, such as unique identifiers and standard information sent by a device for personalised ads and content, ad and content measurement, and audience insights, as well as to develop and improve products. How do I detect unsigned integer overflow? @TheodoreMurdock I think the answer to that question is simple. another point: while compiler would be able to detect this arithmetic condition, some bit operations would be able to fool it so that the condition won't be optimized out. While the historical reason signed overflow was specified as undefined behavior was probably these bogus legacy representations (ones complement/sign-magnitude) and overflow interrupts, the modern reason for it to remain undefined behavior is optimization. To illustrate the use of safely_promote_t, lets write a template function version of Figure 3 that is free from any undefined behavior when T is an unsigned integer type: Of course the best solution of all came from the introductory advice: use a signed integral type instead of unsigned types whenever you can. 8.3.1 Unary operators [expr.unary.op] (parts 7, 8, 10) [For the unary operators +, -, ~, the operands are subject to integral promotion.]. Should I exit and re-enter EU with my EU passport or is it ok? Examples of frauds discovered because someone tried to mimic a random sequence, Counterexamples to differentiation under integral sign, revisited. result that cannot be represented by the resulting unsigned integer value that can be represented by the resulting type. "Overflow" here means "producing a value that doesn't fit the operand". reduced modulo the number that is one greater than the largest value that can be How do I detect unsigned integer overflow? And yet it makes a lot of sense to use from a practical point of view for the particular work I do. Site design / logo 2022 Stack Exchange Inc; user contributions licensed under CC BY-SA. First of all, please note that C11 3.4.3, like all examples and foot notes, is not normative text and therefore not relevant to cite! It overflows. Say we have an 8-bit unsigned value: Unsigned math is clearly defined in C and C++, where signed math is technically either undefined or implementation dependent or some other "things that you wouldn't expect may happen" wording (I don't know the exact wording, but the conclusion is that "you shouldn't rely on the behaviour of overflow in signed integer values"). @AndyRoss: While there are no architectures using anything other than 2s complement (for some definition of no), there. If for our compiler unsigned short is 16 bit and int is 32 bit, then any product of x and y larger than 2^31 will overflow the signed type int. No surprises that theyre designer and advocate of one of the worst languages ever created. So that code like this would be incorrect even if it happens to work on most architectures. an additive inverse. I think your assumption 1) that this can be switched off for any given processor has been false on at least one important historical architecture, the CDC, if my memory is correct. I don't quite follow - why does it help to have an additive inverse? Lets look at a second surprise from unsigned integer promotion: If you run Figure 2 on a system where unsigned short and int are both 16bit types, the program will output sum == 0. :Thanks for the explanations. expression (that is, if the result is not mathematically defined or Where is that defined in the C++ standard? Unsigned arithmetic follow the rules of modulo arithmetic, meaning that 0x0000 - 0x0001 evaluates to 0xFFFF for 32-bit unsigned types. Compiler writers are likely to understand this could break older code that implicitly assumes uint32_t wont be promoted, but theres no guarantee. Another benefit from allowing signed integer overflow to be undefined is that it makes it possible to store and manipulate a variable's value in a processor register that is larger than the size of the variable in the source code. Thus its implementation defined whether inthas a larger bit width than unsigned short, and by extension its implementation defined whetherunsigned shortwill be promoted to typeint. It is true that a signed integer overflow can occur here, which is undefined behavior. Are the S&P 500 and Dow Jones Industrial Average securities? I feel so much better now, knowing that if any unsigned addition rolls around to zero and causes mayhem, it will be because. Why is unsigned integer overflow defined behavior but signed integer overflow isn't? How big can a 64 bit unsigned integer be? Heres what happens: As before, one and max are promoted to type int prior to addition, resulting in a type int summation value of 65536. -fsanitize=vla-bound: A variable-length array whose bound does not evaluate to a positive value. Is the EU Border Guard Agency able to tell Russian passports issued in Ukraine or Georgia from the legitimate ones? MOSFET is getting very hot at high frequency PWM, Connecting three parallel LED strips to the same power supply, confusion between a half wave and a centre tapped full wave rectifier. How can I fix it? Would like to stay longer than 90 days. Not sure if it was just me or something she sent to the whole team. Unsigned Overflow is Well Defined (C99, Section 6.2.5) "A computation involving unsigned operands can never overflow, because a result that cannot be represented by the resulting unsigned integer type is reduced modulo the number that is one greater than the largest value that can be represented by the resulting type." The History This isn't a hard-fast rule, as you'll see near the end, but just how they proceed for unsigned integers. use extra instructions to check for potential overflow and calculate differently in that case). because a result that cannot be represented by the resulting unsigned integer type is int promotion: Is the following well-defined? Web. The variable one will be assigned the value 1, and will retain this value after being converted to type int. http://ptgmedia.pearsoncmg.com/images/0321335724/samplechapter/seacord_ch05.pdf, http://en.wikipedia.org/wiki/Modulo_operation. The historical reason is that most C implementations (compilers) just used whatever overflow behaviour was easiest to implement with the integer representation it used. Efficient unsigned-to-signed cast avoiding implementation-defined behavior. @DavidElliman: It is not only an issue of whether you can detect it, but what the result is. In a sign + value implementation, The important note here, though, is that there remain. a technical reason for this discrepancy? But since integral promotion occurs, the result of a left shift when x is less than y would be undefined behavior. How does 0 flip back to max integer value when subtracting -1? leetcode x x stl vector.h UndefinedBehaviorSanitizer: undefined behavior usr bin .. l What happens when I subtract an unsigned integer from a signed integer in C++? Integer Overflow Risks. @underscore_d Of courseit's clear why they made the design decision. Then you get twos complement for free at the same time. I understand why signed integer overflow is not safe, but it is not the case for value can be represented by the new type, it is unchanged. What juanchopanza said makes sense. Some operations at the machine level can be the same for signed and unsigned numbers. The rubber protection cover does not pass through the hole in the rim. With regard to Figure 4, this means a compiler could assume the conditional (x >= y) in toy_shift() will always succeed because the alternative would be that the function had undefined behavior from left shifting a negative number, and the compiler knows that undefined behavior is impossible for valid code. To learn more, see our tips on writing great answers. 8.6 Multiplicative operators [expr.mul] By clicking Post Your Answer, you agree to our terms of service, privacy policy and cookie policy. Actually, the overflow behavior is always undefined, in theory even for unsigned integers. the number that is one greater than the largest value that can be Signed Integer Overflow Incorrect pairing of memory allocation and deallocation Shifting by an invalid number of positions Multiple non-identical definitions (the One Definition Rule) Modifying a const object Modifying a string literal Accessing an object as the wrong type Overflow during conversion to or from floating point type It doesn't matter if you read factor outside the loop body; if it has overflowed by then then the behaviour of your code on, after, and somewhat paradoxically before the overflow is undefined. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. How does legislative oversight work in Switzerland when there is technically no "opposition" in parliament? However, when interpreting the result of those operations, some cases don't make sense - positive and negative overflow. If he had met some scary fish, he would immediately return to the surface. The other huge example of undefined behavior for the purpose of permitting optimization is the aliasing rules. Is this how the + operator is implemented in C? Should teachers encourage good students to help weaker ones? About Signed: An example of undefined behavior is the behavior on integer overflow. Why does the distance from light to subject affect exposure (inverse square law) while from subject to lens does not? This will cast the temporary and get you the signedness AND truncate the value so you get what you expected. what happens to a typhoon when it makes a landfall. Rust is interesting, but its a chicken and egg problem where I dont want to invest time into something that wont yet have large impact due to few people using it. We do not currently allow content pasted from ChatGPT on Stack Overflow; read our policy here. This sanitizer does not check for lossy implicit conversions performed before such a computation (see -fsanitize=implicit-conversion ). QGIS expression not working in categorized symbology. For example, the C99 standard (6.2.5/9) states > A computation involving unsigned operands can never overow, because a result that cannot be represented by the resulting unsigned integer type is reduced modulo the number that is one greater than the largest value that can be represented by the resulting type. In my opinion, this makes signed integers the odd-one out, not unsigned, but it's fine they offer this fundamental difference as the programmer can still perform well-defined signed operations with overflow. The compiler will implicitly perform integral promotion on line 6, so that the multiplication will involve two (promoted/converted) operands of type int, not of type unsigned short. Well, an unsigned integer subtraction has defined behavior, also it is a tricky thing. If for some reason the platform you're targeting doesn't use 2's Compliment for signed integers, you will pay a small conversion price when casting between uint32 and int32. rev2022.12.11.43106. You can always perform arithmetic operations with well-defined overflow and underflow behavior, where signed integers are your starting point, albeit in a round-about way, by casting to unsigned integer first then back once finished. 0x0000 - 0x0001 == 0x 1 0000 - 0x0001 == 0xFFFF. Why does my stock Samsung Galaxy phone/tablet lack some features compared to other Samsung Galaxy models? Thanks for contributing an answer to Stack Overflow! 1 A prvalue of an integer type other than bool, char16_t, char32_t, or wchar_t whose integer conversion rank (7.15) is less than the rank of int can be converted to a prvalue of type int if int can represent all the values of the source type; otherwise, the source prvalue can be converted to a prvalue of type unsigned int. I have come across code from someone who appears to believe there is a problem subtracting an unsigned integer from another integer of the same type when the result would be negative. Where in the C99 standard does it say that signed integer overflow is undefined behavior? largest value that can be represented by the resulting type. i'm reading an article about integer security . That it can never overflow means that it is not an error situation. repeatedly adding or subtracting one more than the maximum value that Find centralized, trusted content and collaborate around the technologies you use most. Aside from Pascal's good answer (which I'm sure is the main motivation), it is also possible that some processors cause an exception on signed integer overflow, which of course would cause problems if the compiler had to "arrange for another behaviour" (e.g. @phresnel, I understand, this was long time ago, but "no need to explain" is "unspecified behaviour", unlike "undefined" one it produces sane results, which may still differ. As the link says, this is like the modulo operator: http://en.wikipedia.org/wiki/Modulo_operation. That's.. misleading. Good point, but this only means that the C compiler won't help you detect it. unsigned integer addition and undefined behavior in C90. Floating point types of course are never subjected to integral promotion. In that case the undefined behavior will manifest itself as different result (but will not crash!). Lets finally look at a contrived toy function: The subtraction operator in Figure 4 has two unsigned short operands x and y, both of which will be promoted to type int. (Yes, really, gcc does this even at -O0). Because arithmetic modulo is applied, the value always fits the operand, therefore, no overflow. Its up to the compiler to define the exact sizes for the typeschar, unsigned char, signed char, short,unsigned short,int, unsigned int, long, unsigned long, long long, andunsigned long long. The only way to know if one of these types has a larger bit-width than another is to check your compilers documentation, or to compile/run a program that outputs the sizeof() result for the types. They give great advice, but I have mixed feelings on the language and pragmatically sometimes its a good choice and sometimes its not. kfvs 12 weather. Should I give a brutally honest feedback on course evaluations? Surprisingly, all the sized integral types (int32_t, uint64_t, etc) are open to possible integral promotion, dependent upon the implementation-defined size ofint. If either the document doesn't specify what happens under certain conditions or if it simply declares something to be undefined behavior, then it's undefined behavior. A computation involving unsigned operands can never overflow, because a result that cannot be represented by the resulting unsigned integer type is reduced modulo the number that is one greater than the largest value that can be represented by the resulting type. The behavior is then merely implementation-defined, although a signal may be raised. If the sign bit is one, the value shall be modified in one of the following ways: the corresponding value with sign bit 0 is negated (sign and magnitude); the sign bit has the value (2N) (twos complement); the sign bit has the value (2N 1) (ones complement). The loss of transitivity can destroy a tree-type data structure work. I mean you may loss transitivity property (i.e. Undefined, unspecified and implementation-defined behavior. 65536 compares as unequal to sum, since sum was assigned the value 0 earlier. Probably because there is more than one way of representing signed integers. When would I give a checkpoint to my D&D party that they can return to if they die? But to do so, you must cast for it. Why are these constructs using pre and post-increment undefined behavior? We might incorrectly think that the compiler cant make any assumptions about the arguments to the toy_shift() function because it cant predict what arbitrary calling code might do, but the compiler can make some limited predictions. Is there an historical or (even better!) When you want to get the difference between numbers and make sure that the modular arithmetic will not be applied, then you should consider using abs() function defined in stdlib.h: Be very careful, especially while writing conditions, because: The result of a subtraction generating a negative number in an unsigned type is well-defined: As you can see, (unsigned)0 - (unsigned)1 equals -1 modulo UINT_MAX+1, or in other words, UINT_MAX. You can see the provided link for the precise integer promotion rules, but in practice, the rules mean that during a math operation or comparison, any integer types smaller (in bit-width) than type int will be implicitly converted by the compiler to type int. Visualizing the unsigned (0 to max) range with respect to the modulo of max+1 (where max = 2^n): Modulo Addition Rule: (A + B) % C = (A % C + B % C) % C. Thanks for contributing an answer to Stack Overflow! To learn more, see our tips on writing great answers. backwards compatibility has its limits. It's almost always free to cast, and in fact, your compiler might thank you for doing so as it can then optimize on your intentions more aggressively. A computation involving unsigned operands can never overow, because a result that cannot be represented by the resulting unsigned integer type is reduced modulo the number that is one greater than the largest value that can be represented by the resulting type. Does integer overflow cause undefined behavior because of memory corruption? Is energy "equal" to the curvature of spacetime? as opposed to using the implementation dependent signed semantics: 0x0000 - 0x0001 == (unsigned)(0 + -1) == (0xFFFF but also 0xFFFE or 0x8001). Without reviewing the standard, I think overflowing signed values results in undefined behaviour. If I understand it correctly, your examples all assume that you actually want to handle the overflow. For yet another language creator Dennis Ritchie once called C quirky, flawed, and an enormous success. Connect and share knowledge within a single location that is structured and easy to search. Received a 'behavior reminder' from manager. For a structure to be a field, every element of the structure other than the additive identity must have a multiplicative inverse. If the result type is unsigned, then modular arithmetic takes place. The only caveat is that the outcome is implementation defined (not undefined). 8.7 Additive operators [expr.add] Not the answer you're looking for? How to make voltage plus/minus signs bolder? The undefined behaviour of signed arithmetic overflow is used to enable optimisations; for example, the compiler can assume that if a > b then a + 1 > b also; this doesn't hold in unsigned arithmetic where the second check would need to be carried out . the same value in each type is the same. This optimization is new in gcc8. Most C implementations (compilers) just used whatever overflow behaviour was easiest to implement with the integer representation it used. You don't have to use modulo arithmetic to support signed or unsigned values -- personally, I don't use modulo arithmetic when I do my own calculations and I do support unsigned calculations. c++ Share Improve this question Follow Ready to optimize your JavaScript with Rust? Making statements based on opinion; back them up with references or personal experience. It may also allow the compiler to algebraically simplify some expressions (especially those involving multiplication or division) in ways that could give different results than the originally-written order of evaluation if a subexpression contains an overflow, since the compiler is allowed to assume that overflow does not happen with the operands you've given it. @harold: That's a matter of semantics. "strlen(s1) - strlen(s2)" is never less than zero, How to subtract two unsigned ints with wrap around or overflow. So, my idea was revisionist at best. And, 2^31-1 is a Mersenne Prime (but 2^63-1 is not prime). Its the promotion of *unsigned* integral types thats problematic and bug-prone. For example, when multiplying two unsigned short variables a and b, you can write (a+0u)*(b+0u). In the latter case, for example, int8_t result = a - b; (where a and b have int8_t type) you can obtain very weird behavior. Keep in mind that if the subtraction had involved unsigned integral types (as it would appear on the surface), the result would have underflowed in a well-defined manner and wrapped around to become a large positive number, and the left shift would have been well-defined. Is this a clang optimizer bug or an undefined behavior in C? By clicking Post Your Answer, you agree to our terms of service, privacy policy and cookie policy. While the historical reason signed overflow was specified as undefined behavior was probably these bogus legacy representations (ones complement/sign-magnitude) and overflow interrupts, the modern reason for it to remain undefined behavior is optimization. Why does the USA not have a constitutional court? But it does so in a defined way, namely by wrapping in the way they explain. [For scant reassurance, I havent seen a compiler do this (yet) for Figure 4.]. Again, from the C99 standard (3.4.3/1), An example of undened behavior is the behavior on integer overow. Asking for help, clarification, or responding to other answers. But that could never happen because the standard says that unsigned integers don't overflow at all. Avoid using this particular solution on any operand of type uint32_t (or any even larger fixed width type), since unsigned int has an implementation defined size (of at least 16 bits) and this size could be smaller than uint32_t on some systems potentially resulting in an undesired narrowing cast. Infinite loop in a for from 0 to 255 with unsigned char counter. The compiler will cast that result from type int to type unsigned short in order to assign it to variable sum. The real requirement is that unsigned types must have all of their bits participating in the value representation. @DavidElliman Unsigned wraparound on addition is easily detectable (. There are many questions about detection of the integer overflow BEFORE the actual addition/substraction because of possible undefined behavior. It's literally not defined by the standard. How can I prevent the gcc optimizer from producing incorrect bit operations? en.wikipedia.org/wiki/Signed_number_representations. edit: Oh and, "clearly defined" would be "well defined" in C++ parlance :), I think there is (at least) a third one which is something like "implementation detail", but my point was rather that I don't know which level of "it's not certain what happens here" that signed integer math ends up under - does it allow just "strange results" or "anything could happen" (e.g. if there were no unsigned equivalent of the largest integer type, and arithmetic operations on unsigned types behaved as though they were first converted them to larger signed types, then there wouldn't be as much need for defined wrapping behavior, but it's difficult to do calculations in a type which doesn't have e.g. For reference, here are the excerpts/summaries from the relevant parts of the C++17standarddraft: 7.6 Integral promotions [conv.prom] An interesting consequence of the potential for undefined behavior in Figure 4 is that any compiler would be within its rights to generate optimized object code for the function (if the static_assert succeeds) that is very fast and almost certainly unintended by the programmer, equivalent to, To see why, we need to understand how modern compilers can use undefined behavior. Wording: "implementation defined behaviour", when a compiler shall document behaviour, and "undefined behaviour", where compilers can do what they want. What is the difference between const int*, const int * const, and int const *? c++ Integer overflow in spite of using unsigned int and modulo operations. represented by the resulting type. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Im open to change. However, your reasoning about the "signed semantics" in this context is wrong. How disastrous is integer overflow in C++? E.g., incrementing negative numbers is the same that for positive numbers (expect under overflow conditions). even if you evaluate 0 - 1 in the domain of signed type and obtain -1 as the intermediate result, this -1 is still required to produce 0xFFFF when later it gets converted to unsigned type. In other words, before overflow can actually happen, C++ will already have truncated the value. The fact that unsigned integers form a ring (not a field), taking the low-order portion also yields a ring, and performing operations on the whole value and then truncating will behave equivalent to performing the operations on just the lower portion, were IMHO almost certainly considerations. If for our compiler unsigned short is 16 bit and int is 32 bit, then any product of x and y larger than 2^31 will overflow the signed type int. To understand this modular arithmetic, just have a look at these clocks: 9 + 4 = 1 (13 mod 12), so to the other direction it is: 1 - 4 = 9 (-3 mod 12). And unfortunately, signed integral overflow is undefined behavior. If all implementations at that time agreed on what unsigned "overflow" should do, that's a good reason for getting it standardized. Any cpu is twos complement if your compiler simply omits generating the useless signed arithmetic opcodes and uses the unsigned ones for both signed and unsigned types. Connecting three parallel LED strips to the same power supply. How can I use a VPN to access a Russian website that is banned in the EU? No multiplication of values of type unsigned short ever occurs in this function. Integral promotion involves some implementation-defined behavior. Using the terminology in the standard, instead of overflowing the value wraps., @LihO: The only operator in C++ that is context-sensitive and acts differently depending on how its result is used is a custom conversion operator. Otherwise, the new type is signed and the value If x is less than y then the result of the subtraction will be a negative number, and left shifting a negative number is undefined behavior. @AndyRoss but there are still systems (OS + compilers, admittedly with an old history) with one's complement and new releases as of 2013. Also, integer sizes were different back in the day. reduced modulo the number that is one greater than the largest value that can be It can assume that calling code will never use any arguments that result in undefined behavior, because getting undefined behavior would be impossible from valid calling code. Basically problem is you have integer overflow just after multiplication. Is unsigned integer subtraction defined behavior? *PATCH] Remove -fstrict-overflow, default to undefined signed integer and pointer overflow @ 2017-04-26 12:03 Richard Biener 2017-04-26 22:05 ` Eric Botcazou 2017-04-27 16:13 ` Jeff Law 0 siblings, 2 replies; 4+ messages in thread From: Richard Biener @ 2017-04-26 12:03 UTC (permalink / raw It would typically be significantly slower to do this than having hardware support for it, but it's no different from processors that doesn't support floating point in hardware, or similar - it just adds a lot of extra code. It's just like an old-style car odometer. because a result that cannot be represented by the resulting unsigned integer type is A computation involving unsigned operands can never overflow, because a result that cannot be represented by the resulting unsigned integer type is reduced modulo the number that is one greater than the largest value that can be represented by the resulting type. For example, the C99 standard (6.2.5/9) states. Theres no undefined behavior in the program or compiler bugs. usually if you are relying on unsigned overflow, you are using a smaller word width, 8bit or 16bit. Where does the idea of selling dragon parts come from? The most technical reason of all, is simply that trying to capture overflow in an unsigned integer requires more moving parts from you (exception handling) and the processor (exception throwing). what happens if the processor has an overflow trap that fires on integer overflow?). Therefore, x * 10 / 10 == x is a trouble if x * 10 is more than INT_MAX. Making statements based on opinion; back them up with references or personal experience. Note that although it does say "A computation involving unsigned operands can never overflow", which might lead you to believe that it applies only for exceeding the upper limit, this is presented as a motivation for the actual binding part of the sentence: "a result that cannot be represented by the resulting unsigned integer type is In practice, it is only the representations for signed values that may differ according to the implementation: one's complement, two's complement, sign-magnitude. I'd rather handle these cases with. Because correct C++ programs are free of undefined behavior, compilers may produce unexpected results when a program that actually has UB is compiled with optimization enabled: For example, Signed overflow int foo (int x) { return x +1 > x; // either true or UB due to signed overflow } may be compiled as ( demo ) foo (int): movl $ 1, % eax ret Sometimes you really do need unsigned integers. You could run into a few problems with unsigned integer types. Well, the first interpretation is correct. [Binary operators *, /, %]2 E.g., if the maximum is 2^16=32768, then 32760 + 9 = (32760 + 9) % (32768+1) = 0. the new type. Original response edited. Therefore, it optimizes down to checking x > 0. It expects a signed integer, and you provide one. Race conditions are really interesting because they include undefined behavior on hardware . represented by the resulting type. This modulo is applied to results of unsigned-only computations, with the divisor being the maximum value the type can hold. std::abs is not "suitable" for unsigned integers. The C++17 standard has multiple sections that involve integral promotion. The assumption that trapping can be switched off is correct, because, as the implementor of the compiler, you can simply choose never to generate signed-arithmetic opcodes and always use the clean/safe unsigned ones. A computation involving unsigned operands can never overow,because a result that cannot be represented by the resulting unsigned integer type is reduced modulo to the number that is one greater than the largest value that can be represented by the resulting type. A computation involving unsigned operands can never overow,because a confusion between a half wave and a centre tapped full wave rectifier. Site design / logo 2022 Stack Exchange Inc; user contributions licensed under CC BY-SA. A disadvantage is maintainers may not understand its meaning when seeing it. Concentration bounds for martingales with adaptive Gaussian steps. Is unsigned integer subtraction defined behavior? Why is unsigned integer overflow defined behavior but signed integer overflow isn't? Hidden integral promotions and narrowing conversions are subtle, and the results can be surprising, which is usually a very bad thing. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. [ If neither operand has scoped enumeration type, type long double, double, or float,] the integral promotions (7.6) shall be performed on both operands. I was under the impression that this kind of undefined behavior essentially meant that the value of that integer could become unreliable. When you subtract two unsigned integers, result is promoted to higher type int if result (lvalue) type is not specified explicitly. Why is unsigned integer overflow defined behavior but signed integer overflow isn't? This phrase is not restricted to overflow of the upper bound of the type, and applies equally to values too low to be represented. Help us identify new roles for community members, Proposing a Community-Specific Closure Reason for non-English content, Integer addition with overflow on a struct, printing the char variable value using %d. not in the range of representable values for its type), the behavior Unsigned integer overflow is well defined by both the C and C++ standards. Conversion of a negative number to unsigned is defined as yielding the number which, when added to the sign-reversed original number, will yield zero (so converting -5 to unsigned will yield a value which, when added to 5, will yield zero). ], 8.9 Relational operators [expr.rel] For example, the authors of the C++ standard say that it doesn't overflow, because modular arithmetic keeps the result within range; they only use the term to describe signed overflow, which is an error giving undefined behaviour. Is there a clang option to get warnings for signed but not for unsigned integer overflow? How do I put three reasons together in a sentence? Topic archived. This includes doing "the right thing" as well as "calling the police" or "crashing". Asking for help, clarification, or responding to other answers. Your email address will not be published. C intN\u,c,int,undefined-behavior,integer-overflow,C,Int,Undefined Behavior,Integer Overflow,C99int8\u tint16\u t27.18.1.1 They will usually be promoted to typeint during operations and comparisons, and so they will be vulnerable to all the undefined behavior of the signed typeint. @0x499602D2 On most hardware it does, although the compiler can make optimizations which won't. Ready to optimize your JavaScript with Rust? These will promote to signed int at the drop of a hat (C has absolutely insane implicit integer conversion rules, this is one of C's biggest hidden gotcha's), consider: To avoid this, you should always cast to the type you want when you are relying on that type's width, even in the middle of an operation where you think it's unnecessary. By clicking Post Your Answer, you agree to our terms of service, privacy policy and cookie policy. Why it will produce this undefined behavior in the first place? Not being able to have it do either, however, would make it necessary to compare 0003 to 9995, notice that it's less, do the reverse subtraction, subtract that result from 9999, and add 1. 1) A processor that generates exception in this case. implicit conversion of unsigned and signed. If it only reads up to 999,999 miles, then one more mile brings it back to zero. This is useful both for mathematical operations and comparisons to ensure that operands wont get unexpectedly promoted to type int. Browse other questions tagged, Where developers & technologists share private knowledge with coworkers, Reach developers & technologists worldwide. Thanks for contributing an answer to Stack Overflow! The behaviour of int overflow is undefined. Am I missing something? I now see the interpretation I was missing. Find centralized, trusted content and collaborate around the technologies you use most. Concerning unsigned arithmetic, on the other hand, the Standard explicitly specifies that (Paragraph 3.9.1/4): Unsigned integers, declared unsigned , shall obey the laws of arithmetic modulo 2^n where n is the number of bits in the value representation of that particular size of integer Is signed integer overflow undefined behaviour or implementation defined? (ISO/IEC 9899:1999 (E) 6.2.5/9). Connect and share knowledge within a single location that is structured and easy to search. The value 65536 isnt representable in a 16 bit unsigned short (sums type), but the conversion is well-defined in C and C++; the conversion gets performed modulo 2N, where N is the bit width of type unsigned short. In the expression (x + y) > y;, the compiler is allowed to assume that x+ydoesn't overflow (because that would be UB). If a C were to provide a means of declaring a "wrapping signed two's complement" integer, no platform that can run C at all should have much trouble supporting it at least moderately efficiently. Ring-like behaviour flows naturally from that. In addition to the other issues mentioned, having unsigned math wrap makes the unsigned integer types behave as abstract algebraic groups (meaning that, among other things, for any pair of values X and Y, there will exist some other value Z such that X+Z will, if properly cast, equal Y and Y-Z will, if properly cast, equal X). It doesnt matter that overflow of unsigned integral types is well-defined behavior in C and C++. Does a 120cc engine burn 120cc of fuel a minute? Undefined, unspecified and implementation-defined behavior. Are defenders behind an arrow slit attackable? Help us identify new roles for community members, Proposing a Community-Specific Closure Reason for non-English content, weird thing in C: not zero is not equal to one, Strange behaviour when intentionally assigning `i` value greater than INT_MAX, why do integers have different behaviors when they overflow. How do I detect unsigned integer overflow? We do not currently allow content pasted from ChatGPT on Stack Overflow; read our policy here. Having 0003-9995 yield 0008 makes it easy to calculate the latter result. Thank you! "implementation detail" means "it's up to the compiler producer, and there is no need to explain that it is". Following is a code that is sensible to a signed integer overflow. if a > b and b > c it is true that a > c). I know signed overflow is undefined behavior, but wouldn't it too wrap around? The usual arithmetic conversions are performed for operands of arithmetic or enumeration type. Find centralized, trusted content and collaborate around the technologies you use most. Should I exit and re-enter EU with my EU passport or is it ok? Principle: Addition and subtraction of signed integers shall use the same representation as addition and subtraction of unsigned integers Result: -x is represented in the same way as the unsigned number 2B - x, where B is the number of bits in the integer Signed overflow: here be dragons What happens when a signed arithmetic computation overflows? Required fields are marked *. except that there isnt any final narrowing conversion back to unsigned short. Can several CRTs be wired in parallel to one oscilloscope circuit? So is this wrong? For C++, theres a fairly good solution. implementation-defined or an implementation-defined signal is raised. Signed integer overflow has undefined behaviour in C++. a sign bit. Overflow of signed integers has undefined behaviour on overflow because of a number of contributing factors. 2 min read. Tabularray table when is wraped by a tcolorbox spreads inside right margin overrides page borders. Connect and share knowledge within a single location that is structured and easy to search. And unfortunately, signed integral overflow is undefined behavior. Note that unsigned numbers smaller than unsigned int may get promoted to type int before the subtraction, the behavior of a-b will depend upon the size of int. The addition of these two (converted/promoted) type int values results in the value 65536, which is easily representable in a 32 bit int type, and so there wont be any overflow or undefined behavior from the addition. casts from signed -> unsigned int are well defined. A clarification regarding the behavior of unsigned integer types specifically can be found here: The range of nonnegative values of a signed integer type is a subrange http://ptgmedia.pearsoncmg.com/images/0321335724/samplechapter/seacord_ch05.pdf. Why is the federal judiciary of the United States divided into circuits? Why is unsigned integer overflow defined behavior but signed integer overflow isn't? Just because a type is defined to use 2s complement representation, it doesn't follow that arithmetic overflow in that type becomes defined. Is energy "equal" to the curvature of spacetime? In this example, N=16 and thus the conversion of 65536 will result in the value 0, which will be assigned to sum. Why is the federal judiciary of the United States divided into circuits? QGIS expression not working in categorized symbology. Firstly, there are different representations of a signed integer (e.g. In most cases I have encountered, the overflow is undesirable, and you want to prevent it, because the result of a calculation with overflow is not useful. Values stored in unsigned bit-fields and objects of type unsigned char shall be represented using a pure binary notation. C and C++ won't make you pay for that unless you ask for it by using a signed integer. Site design / logo 2022 Stack Exchange Inc; user contributions licensed under CC BY-SA. Is it defined at all? The compiler always assumes that we have written valid code unless it can prove otherwise (in which case wed get a compiler error message). This means that in Figure 1, if the static_assert passes, the assignment. @sleske: It's also very useful for both humans and compilers to be able to apply the associative, distributive, and commutative laws of arithmetic to rewrite expressions and simplify them; for example, if the expression. Thanks for contributing an answer to Stack Overflow! The compiler can therefore conclude that with valid code, there is no scenario in which the conditional could possibly fail, and it could use this knowledge to optimize the function, producing object code that simply returns 0. Which way is not specified in the standard, at least not in C++. This is good and easy advice. Why is Singapore currently considered to be a dictatorial regime and a multi-party democracy by different publications? As I understand it, the original C standard in a large part codified existing practice. Unsigned integer overflow is well defined by both the C and C++ standards. This is the only vaguely relevant quote from the C standard I could find. [==, !=] Why does integer overflow on x86 with GCC cause an infinite loop? unsigned operands can never overflow, because a result that cannot be I suppose one could take that quote to mean that when the right operand is larger the operation is adjusted to be meaningful in the context of modulo truncated numbers. If he had met some scary fish, he would immediately return to the surface. PSE Advent Calendar 2022 (Day 11): The other side of Christmas. Does aliquot matter for final concentration? The result of, or the signal raised by, converting an integer to a signed integer type when the value cannot be . appreciate for reply. In this case I'd rather prefer to write buggy, but simple code. type is converted to another integer type other than _Bool, if the A cast is needed though. Most compilers, when possible, will choose "do the right thing", assuming that is relatively easy to define (in this case, it is). Hmmm, it may not be undefined behavior, because this is conversion of an out-of-range value to a signed integral type, not overflow during calculation. n55 140001 Was the ZX Spectrum used for number crunching? Why the infinite loop when data type is unsigned int? It may also help in situations where it's necessary to detect overflow, since performing calculations and checking whether they overflowed is often easier than checking in advance whether they would overflow, especially if the calculations involve the largest available integer type. can be represented in the new type until the value is in the range of For example, How can I use a VPN to access a Russian website that is banned in the EU? Why is it more safe to place sizeof in malloc first? randomascii . for (unsigned char i = 0; i<=0xff; i++) produces infinite loop. For example, its plausible that there could someday be a compiler that defines intas a 64 bit type, and if so, int32_t and uint32_t will be subject to promotion to that largerinttype. How disastrous is integer overflow in C++? In contrast, the C standard says that signed integer overflow leads to undefined behavior where a program can do anything, including dumping core or overrunning a buffer. , 5 . Computing the Modular Multiplicative Inverse, [C/C++] Surprises and Undefined Behavior From Unsigned Integer Promotion, C/C++ compilerscommonlyuseundefinedbehaviortooptimize. The usual arithmetic conversions are performed on the operands and determine the type of the result. @Andy Ross would you consider "no architectures using anything other than 2's complement " today includes the gamut of DSPs and embedded processors? Furthermore, the overflow results differ depending on the underlying signed representation. Help us identify new roles for community members, Proposing a Community-Specific Closure Reason for non-English content. However, the conditional operator works with operands of type int, and so the right hand side summation never gets a similar conversion down to unsigned short. That means anything is possible including "it worked as I expected". int promotion: Is the following well-defined? some CPUs (DSPs, for example) have saturating arithmetic rather then modulo arithmetic. Are the S&P 500 and Dow Jones Industrial Average securities? For better or worse, modern C/C++ compilerscommonlyuseundefinedbehaviortooptimize, by taking advantage of the fact that undefined behavior is impossible in any valid code. Making statements based on opinion; back them up with references or personal experience. The simple way to test for overflow is to do validation by checking whether the current value is less than the previous value. As far as I can tell, the fact that it's a ring is a consequence, not a cause. Also note that there is an exception if any type is converted to a signed type and the old value can no longer be represented. For example, suppose you had a loop to print the powers of 2: long lng; int n; for (n = 0; n < 34; ++n) { lng = pow (2, n); printf ("%li\n", lng); } Adding overflow checking the way that I described results in this:. Unsigned overflow is well defined per the standard and the compiler also should not be allowed to optimize that comparison away (unless it can determine with absolut certainty an overflow will not happen). And we will see this again for the next step. So, my question is. Why do we use perturbative series if they don't converge? Why do quantum objects slow down when volume increases? @ruslik: sure your mileage may vary. The historical reason is that most C implementations (compilers) just used whatever overflow behaviour was easiest to implement with the integer representation it used. You are right. For example, one operation may treat an integer as an unsigned one and another operation may treat exactly the same integer as a signed one, therefore interpreting the value incorrectly. 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