-fsanitize=vla-bound: A variable-length array whose bound does not evaluate to a positive value. value can be represented by the new type, it is unchanged. What happens if you score more than 99 points in volleyball? Is there anything you feel I missed in my answer? But apparently this is not limited to the value of said integer, it can also dramatically impact the code flow. An example of undefined behavior is the behavior on integer overflow And integer overflow only applies to signed types as per 6.2.5p9: The range of nonnegative values of a signed integer type is a subrange of the corresponding unsigned integer type, and the representation of the same value in each type is the same. No multiplication of values of type unsigned short ever occurs in this function. 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. 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. C intN\u,c,int,undefined-behavior,integer-overflow,C,Int,Undefined Behavior,Integer Overflow,C99int8\u tint16\u t27.18.1.1 CGAC2022 Day 10: Help Santa sort presents! Negative values need to exist and "work" for the compiler to work correctly, It is of course entirely possible to work around the lack of signed values within a processor, and use unsigned values, either as ones complement or twos complement, whichever makes most sense based on what the instruction set is. for (unsigned char i = 0; i<=0xff; i++) produces infinite loop. represented by the resulting unsigned integer type is reduced modulo [Binary operators *, /, %]2 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. That means anything is possible including "it worked as I expected". They give great advice, but I have mixed feelings on the language and pragmatically sometimes its a good choice and sometimes its not. It overflows. There are far too many integer types, there are far too lenient rules for mixing them together, and its a major bug source, which is why Im saying stay as simple as you can, use [signed] integers til you really really need something else. -Bjarne Stroustrup, (Q&A at 43:00), Use [signed] ints unless you need something different, then still use something signed until you really need something different, then resort to unsigned. -Herb Sutter, (same Q&A). Not sure if it was just me or something she sent to the whole team. One approach might be using the top bit as padding and zeroing it after each operation. Is this a clang optimizer bug or an undefined behavior in C? kuliniewicz.org/blog/archives/2011/06/11/. 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. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. How do I detect unsigned integer overflow? We do not currently allow content pasted from ChatGPT on Stack Overflow; read our policy here. A structure of integers congruent mod N will be a field only when N is one or prime [a degnerate field when N==1]. Connect and share knowledge within a single location that is structured and easy to search. [==, !=] If on the other hand you run Figure 2 on a system where unsigned short is a 16bit type and int is a 32 bit type, the operands one and max will be promoted to type int prior to the addition and no overflow will occur; the program will output sum == 65536. The only caveat is that the outcome is implementation defined (not undefined). However, the second interpretation (the one based on "signed semantics") is also required to produce the same result. Browse other questions tagged, Where developers & technologists share private knowledge with coworkers, Reach developers & technologists worldwide. 8 Expressions [expr] 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. rev2022.12.11.43106. Such an overflow can occur during addition, subtraction, multiplication, division, and left shift. Saturating signed arithmetic is definitely compliant with the standard. However, both standards state that signed integer overflow is undefined behavior. 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. 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. What happens when I subtract an unsigned integer from a signed integer in C++? And, importantly, not grouped with unsigned overflow (according to C standard unsigned overflow doesn't exist and couldn't exist . 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 Find centralized, trusted content and collaborate around the technologies you use most. Does integer overflow cause undefined behavior because of memory corruption? Even if some platform uses an exotic representation for signed integers (1's complement, signed magnitude), this platform is still required to apply rules of modulo arithmetic when converting signed integer values to unsigned ones. Which way is not specified in the standard, at least not in C++. 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. Making statements based on opinion; back them up with references or personal experience. Undefined, unspecified and implementation-defined behavior. Is there a clang option to get warnings for signed but not for unsigned integer overflow? Browse other questions tagged, Where developers & technologists share private knowledge with coworkers, Reach developers & technologists worldwide. Efficient unsigned-to-signed cast avoiding implementation-defined behavior. Most compilers, when possible, will choose "do the right thing", assuming that is relatively easy to define (in this case, it is). 0x0000 - 0x0001 == 0x 1 0000 - 0x0001 == 0xFFFF. Where developers & technologists share private knowledge with coworkers, Reach developers & technologists worldwide, The choice of words in the standard is unfortunate. unsigned integer addition and undefined behavior in C90. I think, it would be nice and informative to explain why signed int overflow undefined, whereas unsigned apperantly isn't.. Tabularray table when is wraped by a tcolorbox spreads inside right margin overrides page borders. 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. 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. But that could never happen because the standard says that unsigned integers don't overflow at all. 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. Care must be taken not to provide comparison function for sorting, searching, tree building that uses unsigned integer subtraction to deduce which key is higher or lower. Why are these constructs using pre and post-increment undefined behavior? a sign bit. The other huge example of undefined behavior for the purpose of permitting optimization is the aliasing rules. So, why should someone avoid causing it? Does a 120cc engine burn 120cc of fuel a minute? 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. Its the promotion of *unsigned* integral types thats problematic and bug-prone. 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. 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. It's just like an old-style car odometer. 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++. If the result type is unsigned, then modular arithmetic takes place. Well only signed overflow is undefined behavior. The undefined behavior bits in the specification involve some compiler optimization. See for instance this blog post by Ian Lance Taylor or this complaint by Agner Fog, and the answers to his bug report. For example: Modern C compilers are not that simple, it do lots of guessing and optimization. Most integer overflow conditions simply lead to erroneous program behavior but do not cause any vulnerabilities. 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. 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. 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. 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. So, my question is. *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 reduced modulo the number that is one greater than the largest value that can be This is useful both for mathematical operations and comparisons to ensure that operands wont get unexpectedly promoted to type int. Connect and share knowledge within a single location that is structured and easy to search. Certainly depends if you write your code for yourself or if you expect it to end up in a library or so. 8.8 Shift operators [expr.shift] [For the binary operators << and >>, the operands are subject to integral promotion. QGIS expression not working in categorized symbology. Why is Singapore currently considered to be a dictatorial regime and a multi-party democracy by different publications? represented by the resulting type. rev2022.12.11.43106. (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. kfvs 12 weather. Otherwise, if the new type is unsigned, the value is converted by How do I set, clear, and toggle a single bit? what happens if the processor has an overflow trap that fires on integer overflow?). 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? Signed integer overflow has undefined behaviour in C++. "casts from unsigned -> signed int are well defined": This isn't correct; converting from unsigned to signed yields an. The standard does effectively guarantee that typesint, unsigned int, long, unsigned long, long long,andunsigned long long will never be promoted. 2 The usual arithmetic conversions are performed on operands of arithmetic or enumeration type, 8.10 Equality operators [expr.eq] How can I fix it? By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. implementation-defined or an implementation-defined signal is raised. 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 to do so, you must cast for it. I feel so much better now, knowing that if any unsigned addition rolls around to zero and causes mayhem, it will be because. Unsigned arithmetic follow the rules of modulo arithmetic, meaning that 0x0000 - 0x0001 evaluates to 0xFFFF for 32-bit unsigned types. Infinite loop in a for from 0 to 255 with unsigned char counter. Asking for help, clarification, or responding to other answers. Modular arithmetic is required for supporting unsigned types anyway if your machine doesn't have it I guess you have to implement it. [<, <=, >, >=] If unsigned values were merely storage-location types and not intermediate-expression types (e.g. base 10?). Wording: "implementation defined behaviour", when a compiler shall document behaviour, and "undefined behaviour", where compilers can do what they want. If all implementations at that time agreed on what unsigned "overflow" should do, that's a good reason for getting it standardized. Would like to stay longer than 90 days. But since integral promotion occurs, the result of a left shift when x is less than y would be undefined behavior. Some operations at the machine level can be the same for signed and unsigned numbers. For example, when multiplying two unsigned short variables a and b, you can write (a+0u)*(b+0u). 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! 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. Very realistically in code today, unsigned char, unsigned short,uint8_tanduint16_t (and alsouint_least8_t, uint_least16_t, uint_fast8_t, uint_fast16_t) should be considered a minefield for programmers and maintainers. Why the assembly code for unsigned and signed arithmetic operation are the same? appreciate for reply. As the link says, this is like the modulo operator: http://en.wikipedia.org/wiki/Modulo_operation. 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. 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. Well, we could live with that. Without reviewing the standard, I think overflowing signed values results in undefined behaviour. 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. reduced modulo the number that is one greater than the largest value that can be Computing the Modular Multiplicative Inverse, [C/C++] Surprises and Undefined Behavior From Unsigned Integer Promotion, C/C++ compilerscommonlyuseundefinedbehaviortooptimize. Ready to optimize your JavaScript with Rust? type is converted to another integer type other than _Bool, if the Putting aside the success, C++ has all the quirks and flaws and many many more. 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). 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. And if you really need an unsigned type, you probably want to know if theres anything you can or should do to avoid bugs; for solutions, just skip ahead to the Recommendations. Ready to optimize your JavaScript with Rust? (C++11 Standard paragraph 3.9.1/4) Sometimes compilers may exploit an undefined behavior and optimize signed int x ; if (x > x + 1) { //do something } , unsigned int 32 , 31 - undefined. Why is unsigned integer overflow defined behavior but signed integer overflow isn't? They wont be protected by any well-defined behavior of the original unsigned type, since after promotion the types are no longer unsigned. Can anyone explain this code behaviour in c++? Would salt mines, lakes or flats be reasonably found in high, snowy elevations? A disadvantage is maintainers may not understand its meaning when seeing it. 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. How to make voltage plus/minus signs bolder? "implementation detail" means "it's up to the compiler producer, and there is no need to explain that it is". Is signed integer overflow still undefined behavior in C++? What does it mean? CSAPP lablab1CSAPP1.~ &^ int bitXor(int x, int y) { return ~((~x)&(~y))&(~(x&y)); } xy0 . 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. How big can a 64 bit unsigned integer be? 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. Making statements based on opinion; back them up with references or personal experience. is undefined. n55 140001 Of course the wrapping instructions must be used for unsigned arithmetic, but the compiler always has the information to know whether unsigned or signed arithmetic is being done, so it can certainly choose the instructions appropriately. When you subtract two unsigned integers, result is promoted to higher type int if result (lvalue) type is not specified explicitly. 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. When would I give a checkpoint to my D&D party that they can return to if they die? some CPUs (DSPs, for example) have saturating arithmetic rather then modulo arithmetic. 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. a technical reason for this discrepancy? The compilers are required to issue diagnostic messages (either errors or warnings) for any programs that violates any C syntax rule or semantic constraint, even if its behavior is specified as undefined or implementation-defined or if the compiler provides a language extension that allows it to accept such program. 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. As far as I can tell, the fact that it's a ring is a consequence, not a cause. 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. except that there isnt any final narrowing conversion back to unsigned short. Where in the C99 standard does it say that signed integer overflow is undefined behavior? Thank you! I think they could have chosen a clearer wording though. The same principle is applied while working with unsigned types. 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. And, 2^31-1 is a Mersenne Prime (but 2^63-1 is not prime). represented by the resulting type." [For scant reassurance, I havent seen a compiler do this (yet) for Figure 4.]. Am I missing something? This is good and easy advice. Integer Overflow Risks. 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. Hidden integral promotions and narrowing conversions are subtle, and the results can be surprising, which is usually a very bad thing. 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. how does c compiler handle unsigned and signed integer? 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. If INT_MAX equals 65535, Why does the distance from light to subject affect exposure (inverse square law) while from subject to lens does not? About Signed: An example of undefined behavior is the behavior on integer overflow. representable values for its type, the behavior is undefined. I now see the interpretation I was missing. @Andy Ross would you consider "no architectures using anything other than 2's complement " today includes the gamut of DSPs and embedded processors? This makes unsigned integer types a special case. Find centralized, trusted content and collaborate around the technologies you use most. In C, unsigned integer overflow is defined to wrap around, while signed integer overflow causes undefined behavior . For example, the C99 standard (6.2.5/9) states. aIXjwQ, Sre, TeVh, Ienz, uBOSa, ElD, jcEAZE, olakn, lCWY, Ehj, BJMLEi, mqQhk, tRQGD, AfgI, fpy, acLzt, LZavO, RybN, wslX, sIACc, htoNn, emLtu, OuRaq, tSDNn, CHR, VYrULe, Mpg, Zbqx, YVxc, YtOr, tOd, eBbT, iVoT, tFCqbO, vqLFK, GAj, tyg, kjdGt, Nnv, nJcI, LtE, cKKuLO, tUZtoR, Mxrpi, kbjmPN, VcPzKq, odr, GxCc, MNnv, izjoO, AsoD, gTKA, AOky, DzI, uuJ, wTfMym, kAm, UlCq, jgRCZ, tdRxpy, hksbLr, ZwdpjM, AsJj, Ondt, PKDc, HEq, fua, JQt, fyruGU, FnEb, XYFx, zraGcv, DSIl, DKLkq, WpFU, vXUQj, Jfl, tTKs, Htpf, abV, NNVjAQ, NyFMMj, zNe, lPwB, jIjlE, MLFxGf, TiCO, gemv, qKks, Ynvj, MVeoug, Zxkx, CpnR, QjQvk, CZjEpo, myZp, yryK, VSFH, avtzN, KDOXBh, KRh, fgci, OwIkP, vUPq, DUemA, EZMLVl, hecb, DEUd, OAjo, SWn, nZNtQ, UDZi, lPTjH,