Does a c/c++ compiler optimize constant divisions by power-of-two value into shifts?

Question

Question says it all. Does anyone know if the following...

size_t div(size_t value) {
    const size_t x = 64;
    return value / x;
}

...i

Answer 1

Only when it can determine that the argument is positive. That's the case for your example, but ever since C99 specified round-towards-zero semantics for integer division, it has become harder to optimize division by powers of two into shifts, because they give different results for negative arguments.

In reaction to Michael's comment below, here is one way the division r=x/p;of x by a known power of two p can indeed be translated by the compiler:

if (x<0)
  x += p-1;
r = x >> (log2 p);

Since the OP was asking whether he should think about these things, one possible answer would be "only if you know the dividend's sign better than the compiler or know that it doesn't matter if the result is rounded towards 0 or -∞".