Re: regarding optimization options in phase ordering

[Ian Lance Taylor <iant@google.com>]

pms <pmshiva@gmail.com> writes:

> 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

No code is generated for the statement "b=a".  In effect, the program is
optimized into
  printf("the number is :%d",5);
as that is equivalent to the original program.

This is a very simple compiler optimization.  Please consider taking
future questions about code generation to the mailing list
gcc-help@gcc.gnu.org.  The mailing list gcc@gcc.gnu.org is for gcc
developers.  Thanks.

Ian


>> 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