Question

A very often executed piece of code has the following calculation :

long *lp
char *ep, *cp
...
tlen = (ep - cp) / sizeof (*lp);

Would changing this to:

long *lp
char *ep, *cp
...
tlen = (ep - cp) / sizeof (long);

result in any more efficiency (since the sizeof of calculated at compile time) or would a modern compiler handle this at compile time already. what does gcc do ?

Answer 1

The sizeof operator is always a compile time evaluated construct ⁰, so there is no difference.

The fragment …

 tlen = (ep - cp) / sizeof (*lp);

will therefore be transformed into something not unlike …

 tlen = (ep - cp) / 4;

(assuming that sizeof(long)==4 ¹.), with optimizations applied the next transformation is probably …

 tlen = (ep - cp) >> 2;

More optimizations to come, of course; it’s just a demonstration of a possible consequence of it being a compile time construct ⁰.

I would always prefer "sizeof(_var-name_)" over sizeof(_typename_), as its more generic and doesn’t require manual adjustment when you change the type of the variable (except when you change from array to pointer).

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^{0: Except for variable length arrays.}

^{1: Size differs with platform}