Is signed integer overflow still undefined behavior in C++?

Question

As we know, signed integer overflow is undefined behavior. But there is something interesting in C++11 cstdint documentation:

signed integer type with width of exactly 8, 16, 32 and 64 bits respectively with no padding bits and using 2's complement for negative values (provided only if the implementation directly supports the type)

_{See link}

And here is my question: since the standard says explicitly that for int8_t, int16_t, int32_t and int64_t negative numbers are 2's complement, is still overflow of these types an undefined behavior?

Edit I checked C++11 and C11 Standards and here is what I found:

C++11, §18.4.1:

The header defines all functions, types, and macros the same as 7.20 in the C standard.

C11, §7.20.1.1:

The typedef name intN_t designates a signed integer type with width N, no padding bits, and a two’s complement representation. Thus, int8_t denotes such a signed integer type with a width of exactly 8 bits.

Never forget that the only primary documentation for C++ is the standard. Everything else, even a wiki like CppReference, is a secondary source. That doesn't mean it's wrong; just not entirely reliable. — Nicol Bolas
I would expect it to be UB, there is no exemption for these types in C, I don't see why C++ would add one. — Daniel Fischer
possible duplicate of Do C99 signed integer types defined in stdint.h exhibit well-defined behaviour in case of an overflow? — ecatmur
I'm a bit confused: where is the verb in the sentence "signed integer type with width of exactly 8, 16, 32 and 64 bits respectively with no padding bits and using 2's complement for negative values (provided only if the implementation directly supports the type)?" Is a bit missing? What does it mean? — YSC
C++11 is based on C99, not C11. But this is not important anyway — L. F.

Andy Prowl Andy Prowl · Accepted Answer · 2013-04-24T09:51:44

is still overflow of these types an undefined behavior?

Yes. Per Paragraph 5/4 of the C++11 Standard (regarding any expression in general):

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. [...]

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.

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

This means that the result of an unsigned arithmetic operation is always "mathematically defined", and the result is always within the representable range; therefore, 5/4 does not apply. Footnote 46 explains this:

46) This implies that unsigned arithmetic does not 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 unsigned integer type.

Is signed integer overflow still undefined behavior in C++?

3 Answers