Hm. If you're going to reorder these things, then I would expect either
an error or a warning at that point, because you really do know that a
reference to an uninitialized variable is happening.
We do warn when we see an uninitialized value if -Wuninitialized is on.
We don't warn at every point we make an optimization based on it, nor
do i think we should :)
> also Note that gcc *guarantees* the union trick will work, even though
> the standard does not.
That's good to know, thanks. But frankly that's braindead to require
someone to add all these new union declarations all over their code,
when a simple cast used to suffice, and ultimately the generated code is
the same. And since we have to write code for compilers other than just
gcc, we can't even really rely on the union trick. In this respect, the
standard is broken.
This example is worse, it gives no warning and gives the wrong result
with -O3 -Wstrict-aliasing :
####
#include <stdio.h>
main() {
int i = 0x123456;
int *p = &i;
*(short *)p = 2;
printf("%x\n", i);
}
####
In this case, it's not two different pointers pointing to the same
memory, it's the same pointer. The compiler doesn't even have to guess
whether two different pointers access the same memory - it knows it's
the same pointer,
understand strange results occurring when there's ambiguity, but there
is no ambiguity here.
You are right, there isn't.
We ask the TBAA analyzer "can a store to a short * touch i.
In this case, it says "no", because it's not legal.
