CODEWITHSUNDEEP
c++language-lawyer
Fitzwilliam Bennet-Darcy
Fitzwilliam Bennet-Darcy

Reputation: 1342

Can the multiplication of two unsigned shorts really lead to undefined behaviour?

I have spent some time trawling this site; in particular this question: Is ((a + (b & 255)) & 255) the same as ((a + b) & 255)?

In doing so, I've been led to the conclusion that

int main()
{
    unsigned short i = std::numeric_limits<unsigned short>::max();
    unsigned short j = i;
    auto y = i * j;
}

can lead to undefined behaviour due to a type promotion of i and j to int which subsequently overflows upon the multiplication! Perhaps i and j need not to even be this large.

My conclusion is that, for example, on a system where unsigned short is 16 bits and int is 32 bits, the behaviour can be undefined.

Am I correct here?

Upvotes: 11

Views: 367

Answers (1)

krzaq
krzaq

Reputation: 16431

Yes, it's possible, and your example is likely to be undefined on most desktop architectures.

For the sake of this example, let's assume that int is 32-bit 2's complement type and unsigned short is 16-bit.

I'm using N4140 for the quotes.

Before multiplication, both values are promoted to int:

§ 4.5 [conv.prom] / 1

A prvalue of an integer type other than bool, char16_t, char32_t, or wchar_t whose integer conversion rank (4.13) 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;

Then:

§ 5 [expr] / 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.

Since the result of 65535 * 65535 (4294836225) is not defined in our 32-bit int (with value range [-2147483648,2147483647]), the behaviour is undefined.

Upvotes: 12

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