Jean Delvare wrote:
v = p->field; if (!p) return;
can be seen as equivalent to
if (!p) return; v = p->field;
Heck, no.
You're missing the side-effect of a null pointer dereference crash (for p->field) (even though v is unused before the return). The optimizer is not allowed to make exceptions go away as a result of the hoisting.
I just had to try this out :)
Using gcc 3.3.2 this code sample:
struct test { int code; };
int test_func(struct test *a) { int ret; if (!a) return -1; ret = a->code; return ret; }
is compiled into:
0: 8b 54 24 04 mov 0x4(%esp,1),%edx 4: 83 c8 ff or $0xffffffff,%eax 7: 85 d2 test %edx,%edx 9: 74 02 je d <test_func+0xd> b: 8b 02 mov (%edx),%eax d: c3 ret
whereas this one:
int test_func(struct test *a)
{
int ret;
ret = a->code;
if (!a) return -1;
return ret;
}
is simply compiled into:
0: 8b 44 24 04 mov 0x4(%esp,1),%eax 4: 8b 00 mov (%eax),%eax 6: c3 ret
It seems that gcc is smart enough to know that after we've dereferenced a pointer, if it was NULL, it doesn't matter any more. So it just assumes that if execution reaches that "if" statement then the pointer can not be NULL at all.
So the 2 versions aren't equivalent, and gcc doesn't treat them as such either.
Just a minor nitpick, though: wouldn't it be possible for an application to catch the SIGSEGV and let the code proceed, making invalid the assumption made by gcc?
-- Paulo Marques - www.grupopie.com
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