Does c++ compiler optimize 0*x?

Question

Does c++ compiler optimize 0*x? I mean does it convert to 0 or it actually does the multiplication?

Thanks

Answer 1

It might:

int x = 3;
int k = 0 * 3;
std::cout << k;

00291000  mov         ecx,dword ptr [__imp_std::cout (29203Ch)] 
00291006  push        0    
00291008  call        dword ptr [__imp_std::basic_ostream<char,std::char_traits<char> >::operator<< (292038h)] 

It even optimizes away the variables altogether.

But it might not:

struct X
{
    friend void operator *(int first, const X& second)
    {
       std::cout << "HaHa! Fooled the optimizer!";
    }
};

//...
X x;
0 * x;

Answer 2

If x is a primitive integral type than the code generator will use optimizations generally referred to as "Arithmetic Rules" to make changes such as:

int x = ...;
y = 0 * x;   ===> y = 0
y = 1 * x;   ===> y = x
y = 2 * x;   ===> y = x + x;

but only for integral types.

If x is of a non-integral type than 0 * x might not always be equal to 0, or may have side-effects.