gcc return struct code generation

[jose gomez valcarcel]

Hi,

You are right and this is the reason. The x86_64 ABI explicity mention this.

The definitive proof is the generated code for:

extern long calc (long);

long fn (long m)
{
    return calc (m) * 2;
}

0000000000000000 <fn>:
   0:48 83 ec 08          sub    $0x8,%rsp
   4:e8 00 00 00 00       callq  9 <fn+0x9>
   9:48 83 c4 08          add    $0x8,%rsp
   d:48 01 c0             add    %rax,%rax
  10:c3                   retq   

The code generated is a little more clever to me. The code with return a struct is also performing a 16 byte align substracting 40 bytes.

You have to told me this two times. Excuse me for the inconvenience.

Regards.


----- Mensaje original -----
De: Ian Lance Taylor <iant@google.com>
Para: jose gomez valcarcel <jcgv33@yahoo.es>
CC: "gcc-help@gcc.gnu.org" <gcc-help@gcc.gnu.org>
Enviado: martes 30 de agosto de 2011 19:45
Asunto: Re: gcc return struct code generation

jose gomez valcarcel <jcgv33@yahoo.es> writes:

> For curiosity, today i test this code with the intel compiler (12.0.5.220). It generates this code for fn:
>
> icc -O2 -fomit-frame-pointer -c z.c
>
>
> 0000000000000000 <fn>:
>    0:56                   push   %rsi
>    1:e8 00 00 00 00       callq  6 <fn+0x6>
>    6:48 03 c2             add    %rdx,%rax
>    9:59                   pop    %rcx
>    a:c3                   retq   
>
> Yo can see the code changes the values of registers RSI and RCX with apparently no reason ! (appart of computing correctly the c code of fn).
> Also this code uses stack with apparently no reason, or i can not see it.

The stack has to be aligned at a 16-byte boundary before the function
call.  The function call itself pushes 8 bytes onto the stack.  So
another 8 byte adjustment is required before making another function
call.  icc is choosing to do that adjustment using push and pop.  I
don't know why that is preferable to adding and subtracting a constant,
but there may well be a reason.

Ian