On 07/20/2018 06:19 AM, Umesh Kalappa wrote:
Hi All ,
We are looking at the C sample i.e
extern int i,j;
int test()
{
while(1)
{ i++;
j=20;
}
return 0;
}
command used :(gcc 8.1.0)
gcc -S test.c -O2
the generated asm for x86
.L2:
jmp .L2
we understand that,the infinite loop is not deterministic ,compiler
is free to treat as that as UB and do aggressive optimization ,but we
need keep the side effects like j=20 untouched by optimization .
Please note that using the volatile qualifier for i and j or empty
asm("") in the while loop,will stop the optimizer ,but we don't want
do that.
Anyone from the community ,please share their insights why above
transformation is right ?
The loop isn't necessarily undefined (and compilers don't look
for undefined behavior as opportunities to optimize code), but
because it doesn't terminate it's not possible for a conforming
C program to detect the side-effects in its body. The only way
to detect it is to examine the object code as you did.
Compilers are allowed (and expected) to transform source code
into efficient object code as long as the transformations don't
change the observable effects of the program. That's just what
happens in this case.
Martin