Quoting "H.J. Lu" <hjl.to...@gmail.com>:
On Sun, May 13, 2012 at 6:17 PM, <amyl...@spamcop.net> wrote:
Quoting "H.J. Lu" <hjl.to...@gmail.com>:
What is the expect run-time behavior when a + b has
overflow/underflow?
The expectation is wrap-around. Note that loop strenght reduction can
cause assumed wrap-around semantics in RTL for strictly conforming C input
where no such wrap-around is in evidence.
I noticed that also. But my impression is loop strength reduction doesn't
use wrap-around address for load/store directly.
I've actually seen it for loop strength reduction, but here is
an example that does not even involve loop strength reduction to
get into trouble - it just involves the distributive law in the
indexed access itself:
extern int a[];
void f (int o)
{
int i;
for (i = C; i < C + 100; i++)
{
a[o-i] = 0;
}
}
At -O2, gcc (GCC) 4.7.0 20120504 (Red Hat 4.7.0-4) for i686 gives:
f:
.LFB0:
.cfi_startproc
movl 4(%esp), %ecx
movl $100, %eax
.p2align 4,,7
.p2align 3
.L2:
leal (%eax,%ecx), %edx
subl $1, %eax
movl $0, a-1200000400(,%edx,4)
jne .L2
rep
ret
.cfi_endproc
Now consider what happens if o == C, and a is within the first GB.
unless the base address is encoded as 64 bit, you'll have an overflow.
