C++ Arithmetic With Mixed Integral Types That Causes Overflow

Question

I have done some tests in VC++2010 mixing operands of different sizes that cause overflow in add operation:

int _tmain(int argc, _TCHAR* argv[])
{
    __int8 a=127;
    __int8 b=1;
    __int16 c=b+a;
    __int8  d=b+a;
    printf("c=%d,d=%d\n",c,d);

    return 0;
}

//result is: c=128, d=-128

I don't understand why c==128! My understanding is that in both additions, b+a are still considered addition of 2 signed 8 bit variables. So the result is an overflow i.e. -128. After that, the result is then promoted to 16 bit signed int for the first assignment operation and c should still get a 16 bit -128 value. Is my understanding correct? The c++ standard is a bit difficult to read. Chapter 4 seems talking about integeral promotion but I can't find anything related to this specific example.

Answer 1

My understanding is that in both additions, b+a are still considered addition of 2 signed 8 bit variables. So the result is an overflow i.e. -128.

No, the promotion happens before the + is evaluated, not after it. The addition happens when both a and b are positive. Both numbers are promoted to ints for an addition, added as two positive numbers, and then converted to a 16-bit short. At no point in the process does the result become negative due to an overflow, hence the end result of 128.

Arguably, this makes sense: the behavior of a and b matches that of two numbers in mathematics, making it more intuitive to language practitioners.