is unsigned integer overflow undefined behaviour

Why is the federal judiciary of the United States divided into circuits? randomascii . What happens if you score more than 99 points in volleyball? kfvs 12 weather. How do I detect unsigned integer overflow? Asking for help, clarification, or responding to other answers. By clicking Post Your Answer, you agree to our terms of service, privacy policy and cookie policy. So, my question is. Two's complement representation allows certain operations to make more sense in binary format. @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. 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 How disastrous is integer overflow in C++? Making statements based on opinion; back them up with references or personal experience. But that could never happen because the standard says that unsigned integers don't overflow at all. So, my idea was revisionist at best. 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. One approach might be using the top bit as padding and zeroing it after each operation. 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. [] 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. i'm reading an article about integer security . They wont be protected by any well-defined behavior of the original unsigned type, since after promotion the types are no longer unsigned. Topic archived. (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. Signed integer overflow has undefined behaviour in C++. 1 The additive [binary] operators + and group left-to-right. It's literally not defined by the standard. Connect and share knowledge within a single location that is structured and easy to search. Theres no undefined behavior in the program or compiler bugs. the new type. I.e. casts from signed -> unsigned int are well defined. 0x0000 - 0x0001 == 0x 1 0000 - 0x0001 == 0xFFFF. Otherwise, if the new type is unsigned, the value is converted by How can I use a VPN to access a Russian website that is banned in the EU? Unsigned integer overflow is well defined by both the C and C++ standards. 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"? 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. Why is unsigned integer overflow defined behavior but signed integer overflow isn't? Connect and share knowledge within a single location that is structured and easy to search. Why is unsigned integer overflow defined behavior but signed integer overflow isn't? In that case the undefined behavior will manifest itself as different result (but will not crash!). 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:. confusion between a half wave and a centre tapped full wave rectifier. So it would be implementation-defined instead. Infinite loop in a for from 0 to 255 with unsigned char counter. 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. is undefined. , unsigned int 32 , 31 - undefined. And here "anything can happen" means "2's complement arithmetic". Ready to optimize your JavaScript with Rust? Thanks for contributing an answer to Stack Overflow! I think you mean 16 bit unsigned types, not 32 bit. representable values for its type, the behavior is undefined. A structure of integers congruent mod N will be a field only when N is one or prime [a degnerate field when N==1]. 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. It's just amusing that they wrote the spec roughly as "there is no arithmetic over/underflow because the data type is spec'd as a ring", as if this design choice meant that programmers don't have to carefully avoid over- and under-flow or have their programs fail spectacularly. However, when interpreting the result of those operations, some cases don't make sense - positive and negative overflow. Is energy "equal" to the curvature of spacetime? 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. In my eyes this is more of an oddity in the C spec. In the expression (x + y) > y;, the compiler is allowed to assume that x+ydoesn't overflow (because that would be UB). 8.8 Shift operators [expr.shift] [For the binary operators << and >>, the operands are subject to integral promotion. Apple doesn't understand integer overflow, recommends undefined behavior. And, importantly, not grouped with unsigned overflow (according to C standard unsigned overflow doesn't exist and couldn't exist . To subscribe to this RSS feed, copy and paste this URL into your RSS reader. If the result type is unsigned, then modular arithmetic takes place. casts from unsigned -> signed int are well defined. 11 Many binary operators that expect operands of arithmetic or enumeration type cause conversions [These are] called the usual arithmetic conversions. How can I use a VPN to access a Russian website that is banned in the EU? Asking for help, clarification, or responding to other answers. reduced modulo the number that is one greater than the largest value that can be QGIS expression not working in categorized symbology. 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. For example, when multiplying two unsigned short variables a and b, you can write (a+0u)*(b+0u). If I decrement `std::size_t(0)` is that guaranteed to be equal to `std::size_t(-1)`? Casts between signed and unsigned integer types of the same width are free, if the CPU is using 2's compliment (nearly all do). appreciate for reply. 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. std::abs is not "suitable" for unsigned integers. It doesnt matter that overflow of unsigned integral types is well-defined behavior in C and C++. Should I give a brutally honest feedback on course evaluations? And, 2^31-1 is a Mersenne Prime (but 2^63-1 is not prime). 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. For example, To learn more, see our tips on writing great answers. Are the S&P 500 and Dow Jones Industrial Average securities? "strlen(s1) - strlen(s2)" is never less than zero, How to subtract two unsigned ints with wrap around or overflow. That's.. misleading. Is this how the + operator is implemented in C? This helps in situations where wrap-around behavior is actually useful - for example with TCP sequence numbers or certain algorithms, such as hash calculation. This is the only vaguely relevant quote from the C standard I could find. If signed integer arithmetic overflows, it results in undefined behaviour. I think they could have chosen a clearer wording though. Is this a clang optimizer bug or an undefined behavior in C? I'd rather handle these cases with. Though if youre curious, you probably also want to know why its good advice. 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. 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. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. 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. usually if you are relying on unsigned overflow, you are using a smaller word width, 8bit or 16bit. In Figure 1, the unsigned short variable max will be assigned the value 65535, and will retain this value when converted to type int. Since 0u has type unsigned int, adding it to a or b will effectively manually promote the operand to unsigned int if it has type smaller than unsigned int. Signed overflow is Undefined Behaviour in ISO C. You can't reliably cause it and thencheck if it happened. @AndyRoss: While there are no architectures using anything other than 2s complement (for some definition of no), there. When you subtract two unsigned integers, result is promoted to higher type int if result (lvalue) type is not specified explicitly. Not the answer you're looking for? So, why should someone avoid causing it? [Binary operators *, /, %]2 How does the Integer addition result arrive at its value in the case of overflow in C, 64-bit Multiplication Overflow Gets 0 in C++. 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. This helper class provides a safe and relatively easy way to achieve well-defined behavior with all unsigned integer types, as well see by example. The explicit casting prevents implicit promotion of unsigned types to int. 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. The simple way to test for overflow is to do validation by checking whether the current value is less than the previous value. (ISO/IEC 9899:1999 (E) 6.2.5/9) 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. Your email address will not be published. But this leaves far more integral types than you might expect which may (at least in principle) be promoted. rev2022.12.11.43106. How disastrous is integer overflow in C++? And unfortunately, signed integral overflow is undefined behavior. What is this fallacy: Perfection is impossible, therefore imperfection should be overlooked. :Thanks for the explanations. 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. But apparently this is not limited to the value of said integer, it can also dramatically impact the code flow. If unsigned values were merely storage-location types and not intermediate-expression types (e.g. 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. It's not just different machines; it's different optimization levels within the same compiler on the same machine: @R.. You could keep a sign bit. Something can be done or not a fit? Since unsigned short and int are the same size, the operands one and max will not be promoted, and the addition will overflow in a well defined manner resulting in 0. About Signed: An example of undefined behavior is the behavior on integer overflow. 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. Is this a clang optimizer bug or an undefined behavior in C? Making statements based on opinion; back them up with references or personal experience. of the corresponding unsigned integer type, and the representation of 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. Should I exit and re-enter EU with my EU passport or is it ok? Does integer overflow cause undefined behavior because of memory corruption? This makes unsigned integer types a special case. value that can be represented by the resulting type. It is true that a signed integer overflow can occur here, which is undefined behavior. some CPUs (DSPs, for example) have saturating arithmetic rather then modulo arithmetic. For example, the C99 standard (6.2.5/9) states. 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. How to make voltage plus/minus signs bolder? You are right. represented by the resulting type." what happens if the processor has an overflow trap that fires on integer overflow?). How do I arrange multiple quotations (each with multiple lines) vertically (with a line through the center) so that they're side-by-side? Hidden integral promotions and narrowing conversions are subtle, and the results can be surprising, which is usually a very bad thing. Compiler writers are likely to understand this could break older code that implicitly assumes uint32_t wont be promoted, but theres no guarantee. Undefined, unspecified and implementation-defined behavior. 65536 compares as unequal to sum, since sum was assigned the value 0 earlier. The C++17 standard has multiple sections that involve integral promotion. You could run into a few problems with unsigned integer types. Find centralized, trusted content and collaborate around the technologies you use most. How big can a 64 bit unsigned integer be? An example: OS 2200. The loss of transitivity can destroy a tree-type data structure work. The usual arithmetic conversions are performed for operands of arithmetic or enumeration type. Unlike signed integer overflow, this is not undefined behavior, but it is often unintentional. Undefined, unspecified and implementation-defined behavior. Firstly, there are different representations of a signed integer (e.g. This phrase is not restricted to overflow of the upper bound of the type, and applies equally to values too low to be represented. 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. 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. 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. . 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. Where in the C99 standard does it say that signed integer overflow is undefined behavior? @ruslik: sure your mileage may vary. Again, from the C99 standard (3.4.3/1), An example of undened behavior is the behavior on integer overow. So that code like this would be incorrect even if it happens to work on most architectures. E.g., incrementing negative numbers is the same that for positive numbers (expect under overflow conditions). result that cannot be represented by the resulting unsigned integer Is energy "equal" to the curvature of spacetime? Certainly depends if you write your code for yourself or if you expect it to end up in a library or so. QGIS expression not working in categorized symbology. And yet it makes a lot of sense to use from a practical point of view for the particular work I do. When would I give a checkpoint to my D&D party that they can return to if they die? Where developers & technologists share private knowledge with coworkers, Reach developers & technologists worldwide, The choice of words in the standard is unfortunate. 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. Tabularray table when is wraped by a tcolorbox spreads inside right margin overrides page borders. not in the range of representable values for its type), the behavior What does it mean? Basically problem is you have integer overflow just after multiplication. 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. The variable one will be assigned the value 1, and will retain this value after being converted to type int. So is this wrong? implicit conversion of unsigned and signed. Unsigned numbers are simply integers represented using binary digits instead of decimal digits. because a result that cannot be represented by the resulting unsigned integer type is If it only reads up to 999,999 miles, then one more mile brings it back to zero. 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. "implementation detail" means "it's up to the compiler producer, and there is no need to explain that it is". The rubber protection cover does not pass through the hole in the rim. An alternative you can use is to explicitly cast an operand to unsigned int, which works fine for unsigned char, unsigned short, uint8_t and uint16_t. 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. It is also worth noting that "undefined behaviour" doesn't mean "doesn't work". 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. The standard does effectively guarantee that typesint, unsigned int, long, unsigned long, long long,andunsigned long long will never be promoted. 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. largest value that can be represented by the resulting type. Ready to optimize your JavaScript with Rust? base 10?). Probably because there is more than one way of representing signed integers. The problem is that signed integer overflow causes undefined behaviour. A computation involving unsigned operands can never overow,because a Would salt mines, lakes or flats be reasonably found in high, snowy elevations? Is signed integer overflow still undefined behavior in C++? How many transistors at minimum do you need to build a general-purpose computer? Why do quantum objects slow down when volume increases? C implementations usually used the same representation used by the CPU - so the overflow behavior followed from the integer representation used by the CPU. Why is it more safe to place sizeof in malloc first? Connect and share knowledge within a single location that is structured and easy to search. The behavior is then merely implementation-defined, although a signal may be raised. I got confused by the prime power moduli. Otherwise, the new type is signed and the value @R: 2's complement isn't the only issue though - e.g. How does legislative oversight work in Switzerland when there is technically no "opposition" in parliament? Luckily, C2x will solve the `uint8_t` and `uint16_t` problem with `_BitInt(8)` and `_BitInt(16)`, which dont promote to `int` automatically. use integer type which can hold a result of such magnitude, for example: unsigned long long int 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. 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 does 0 flip back to max integer value when subtracting -1? [] A computation involving unsigned operands can never overflow, How do I set, clear, and toggle a single bit? 8 Expressions [expr] can be represented in the new type until the value is in the range of 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. 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 same principle is applied while working with unsigned types. repeatedly adding or subtracting one more than the maximum value that Which way is not specified in the standard, at least not in C++. -fsanitize=vla-bound: A variable-length array whose bound does not evaluate to a positive value. I was under the impression that this kind of undefined behavior essentially meant that the value of that integer could become unreliable. A disadvantage is maintainers may not understand its meaning when seeing it. As the link says, this is like the modulo operator: http://en.wikipedia.org/wiki/Modulo_operation. 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. Methods to address integer overflow problems [ edit] Detection [ edit] Run-time overflow detection implementation UBSan ( undefined behavior sanitizer) is available for C compilers . See for instance this blog post by Ian Lance Taylor or this complaint by Agner Fog, and the answers to his bug report. Well only signed overflow is undefined behavior. Why does the USA not have a constitutional court? 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. a sign bit. http://ptgmedia.pearsoncmg.com/images/0321335724/samplechapter/seacord_ch05.pdf, http://en.wikipedia.org/wiki/Modulo_operation. How many transistors at minimum do you need to build a general-purpose computer? Asking for help, clarification, or responding to other answers. I was watching this video where integer overflow is mentioned. Connect and share knowledge within a single location that is structured and easy to search. What's your point ? Not sure if it was just me or something she sent to the whole team. Putting aside the success, C++ has all the quirks and flaws and many many more. Is unsigned integer subtraction defined behavior? Having 0003-9995 yield 0008 makes it easy to calculate the latter result. As J-16 SDiZ hinted at, the fact that signed overflow is undefined behavior allows the compiler to optimize out some conditionals whose algebraic truth (but not necessarily representation-level truth) are already established by a previous branch. What juanchopanza said makes sense. 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. represented by the resulting type. For a structure to be a field, every element of the structure other than the additive identity must have a multiplicative inverse. 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 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). how does c compiler handle unsigned and signed integer? And we will see this again for the next step. Values stored in unsigned bit-fields and objects of type unsigned char shall be represented using a pure binary notation. That's what makes it undefined. int a = UINT_MAX;is not an instance of signed integer overflow, this definition involves a conversion from unsigned intto intwith a value that exceeds the range of type int. Im open to change. Would like to stay longer than 90 days. The answer depends upon the implementation of the compiler. No surprises that theyre designer and advocate of one of the worst languages ever created. I think, it would be nice and informative to explain why signed int overflow undefined, whereas unsigned apperantly isn't.. Following is a code that is sensible to a signed integer overflow. Is this a clang optimizer bug or an undefined behavior in C? Making statements based on opinion; back them up with references or personal experience. And unfortunately, signed integral overflow is undefined behavior. Which or what interpretation is right? an additive inverse. c++ Share Improve this question Follow This isn't a hard-fast rule, as you'll see near the end, but just how they proceed for unsigned integers. Should I exit and re-enter EU with my EU passport or is it ok? 2 min read. [==, !=] @harold It is from n1570 standard 6.2.5/9. If you want such behaviour from other types, then do your arithmetic followed by applying the required modulus; that uses fundamental operators. 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. 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. This modulo is applied to results of unsigned-only computations, with the divisor being the maximum value the type can hold. 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. type is converted to another integer type other than _Bool, if the 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. Browse other questions tagged, Where developers & technologists share private knowledge with coworkers, Reach developers & technologists worldwide. [ If neither operand has scoped enumeration type, type long double, double, or float,] the integral promotions (7.6) shall be performed on both operands. By clicking Post Your Answer, you agree to our terms of service, privacy policy and cookie policy. 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. 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. There are many questions about detection of the integer overflow BEFORE the actual addition/substraction because of possible undefined behavior. I know signed overflow is undefined behavior, but wouldn't it too wrap around? 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. The misbehavior can even precede the overflow. Can several CRTs be wired in parallel to one oscilloscope circuit? "implementation detail" would be "implementation defined" in C++ parlance. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. 8.11 Bitwise AND operator [expr.bit.and]1 The usual arithmetic conversions are performed; 8.12 Bitwise exclusive OR operator [expr.xor]1 The usual arithmetic conversions are performed; 8.13 Bitwise inclusive OR operator [expr.or]1 The usual arithmetic conversions are performed; Bjarne Stroustrup and Herb Sutter both give absolutely awful advice. How do I put three reasons together in a sentence? Not the answer you're looking for? Compilers do not want you to rely on them doing the right thing, though, and most of them will show you so as soon as you compile, For an example program that gives different results when faced with. integer promotion and unsigned interpretation. The historical reason is that most C implementations (compilers) just used whatever overflow behaviour was easiest to implement with the integer representation it used. Examples of frauds discovered because someone tried to mimic a random sequence, Counterexamples to differentiation under integral sign, revisited. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. 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? We do not currently allow content pasted from ChatGPT on Stack Overflow; read our policy here. 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. Integer Overflow Risks. Site design / logo 2022 Stack Exchange Inc; user contributions licensed under CC BY-SA. If he had met some scary fish, he would immediately return to the surface. But it does so in a defined way, namely by wrapping in the way they explain. One more example to show unsigned data type wraps around instead of overflow: Assigning a -ve number to the unsigned is not recommended but for the illustrative purpose, I'm using it below. It's just like an old-style car odometer. Where does the idea of selling dragon parts come from? 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). Surprisingly, all the sized integral types (int32_t, uint64_t, etc) are open to possible integral promotion, dependent upon the implementation-defined size ofint. 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. How can I prevent the gcc optimizer from producing incorrect bit operations? Find centralized, trusted content and collaborate around the technologies you use most. rev2022.12.11.43106. But to do so, you must cast for it. How do I detect unsigned integer overflow? The integral promotion results in non-portable code. Overflow of signed integers has undefined behaviour on overflow because of a number of contributing factors. No multiplication of values of type unsigned short ever occurs in this function. 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. To learn more, see our tips on writing great answers. They didn't agree on what signed overflow should do, so that did not get in the standard. use extra instructions to check for potential overflow and calculate differently in that case). , 5 . Why does my stock Samsung Galaxy phone/tablet lack some features compared to other Samsung Galaxy models? as opposed to using the implementation dependent signed semantics: 0x0000 - 0x0001 == (unsigned)(0 + -1) == (0xFFFF but also 0xFFFE or 0x8001). 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