Shankar Unni wrote:
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
All that is necessary for the triumph of evil is that good men do nothing.
Edmund Burke (1729 - 1797)
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