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168

answers:

5

Edited to include proper standard reference thanks to Carl Norum.

The C standard states

If an exceptional condition occurs during the evaluation of an expression (that is, if the result is not mathematically defined or not in the range of representable values for its type), the behavior is undefined.

Are there compiler switches that guarantee certain behaviors on integer overflow? I'd like to avoid nasal demons. In particular, I'd like to force the compiler to wrap on overflow.

For the sake of uniqueness, let's take the standard to be C99 and the compiler to be gcc. But I would be interested in answers for other compilers (icc, cl) and other standards (C1x, C89). In fact, just to annoy the C/C++ crowd, I'd even appreciate answers for C++0x, C++03, and C++98.

Note: International standard ISO/IEC 10967-1 may be relevant here, but as far as I could tell it was mentioned only in the informative annex.

A: 

I'm not sure if there are any compiler switches you can use to enforce uniform behavior for overflows in C/C++. Another option is to use the SafeInt<T> template. It's a cross platform C++ template that provides definitive overflow / underflow checks for all types of integer operations.

JaredPar
+5  A: 

For your C99 answer, I think 6.5 Expressions, paragraph 5 is what you're looking for:

If an exceptional condition occurs during the evaluation of an expression (that is, if the result is not mathematically defined or not in the range of representable values for its type), the behavior is undefined.

That means if you get an overflow, you're out of luck - no behaviour of any kind guaranteed. Unsigned types are a special case, and never overflow (6.2.5 Types, paragraph 9):

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.

C++ has the same statements, worded a bit differently:

  • 5 Expressions, paragraph 4:

    If during the evaluation of an expression, the result is not mathematically defined or not in the range of representable values for its type, the behavior is undefined. [Note: most existing implementations of C++ ignore integer overflows. Treatment of division by zero, forming a remainder using a zero divisor, and all floating point exceptions vary among machines, and is usually adjustable by a library function. —endnote]

  • 3.9.1 Fundamental types, paragraph 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.

Carl Norum
Yes, that's what I was looking for. (Though the reference confused me at first -- that's 6.5 paragraph 5, not section 6.5.5.) Do you know of any way to avoid this? Wrapping on overflow is common, and there are many times when I want this to happen. Do any popular compilers have a switch that makes them promise to wrap?
Charles
@Charles, undefined behaviour is undefined. You may get lucky with your particular compiler - check its documentation for a statement that will give you peace of mind. In `gcc`, for example, you can check out the `-fstrict-overflow` and `-fwrapv` flags.
Carl Norum
+4  A: 

In C99 the general behavior is desribed in 6.5/5

If an exceptional condition occurs during the evaluation of an expression (that is, if the result is not mathematically defined or not in the range of representable values for its type), the behavior is undefined.

The behavior of unsigned types is described in 6.2.5/9, which basically states that operations on unsigned types never lead to exceptional condition

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.

GCC compiler has a special option -ftrapv, which is intended to catch run-time overflow of signed integer operations.

AndreyT
+5  A: 

Take a look at -ftrapv and -fwrapv:

-ftrapv

This option generates traps for signed overflow on addition, subtraction, multiplication operations.

-fwrapv

This option instructs the compiler to assume that signed arithmetic overflow of addition, subtraction and multiplication wraps around using twos-complement representation. This flag enables some optimizations and disables other. This option is enabled by default for the Java front-end, as required by the Java language specification.

Matt Joiner
Excellent, just what I wanted to see. Is there anything comparable for unsigned types?
Charles
@Charles you don't need them for unsigned types - overflow behaviour is already well-defined for them (see my answer).
Carl Norum
@Carl Norum: I see that it's defined in C++, and that it's defined in C for shifts (6.5.7 para 5). I can't see where it's defined in the C standard for addition and multiplication.
Charles
@Charles - 6.2.5 paragraph 9: "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."
Carl Norum
Note that `-fwrapv` does **not** guarantee that overflow will wrap. All it does is tell the optimiser that it can *assume* that that's the case. Whether it really is true depends on your machine architecture.
caf
+1  A: 

6.2.5 paragraph 9 is what you're looking for:

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.31) 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.

Chris Dodd