Thanx Ian,
My question is as follwos
We've a problem here. we were trying to use cc1 with & without -O option to
verify the optimizations happening in our sample code.
the sample code is given below
file name : 1.c
#include
int main()
{ int a=5;
int b;
b=a;
printf("the number is :%d",b); }
Here, in 1.c.026t.copyrename1, we get the following output
1.c.026t.copyrename1
;; Function main (main)
main ()
{ int b; int a; : a_2 = b_1(D); return; }
but in 1.c.027t.ccp1, the output does not contain the actual assignment
b=a.
;; Function main (main)
main ()
{ int b; int a; : return; }
We want to know, without b=a, how is it generating the following final code
for b=a
pushl %ebp
> movl %esp, %ebp
> andl $-16, %esp
> subl $16, %esp
> movl $5, 4(%esp)
> movl $.LC0, (%esp)
> call printf
Kindly help
> thanks, But b=a is a assignment statement. It is doing some memory
> operations
> isn't it. Assuming b=a is a dead statement, how r the following i386
> assembly statements generated
> pushl %ebp
> movl %esp, %ebp
> andl $-16, %esp
> subl $16, %esp
> movl $5, 4(%esp)
> movl $.LC0, (%esp)
> call printf
The first four statements set up the stack. The last three do the
printf statement.
What is your real question?
Ian
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