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From: "d.salikhov at samsung dot com" <gcc-bugzilla@gcc.gnu.org>
To: gcc-bugs@gcc.gnu.org
Subject: [Bug target/67797] New: [ARM] Unnecessary r0 saving for memset call
Date: Thu, 01 Oct 2015 14:19:00 -0000
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https://gcc.gnu.org/bugzilla/show_bug.cgi?id=67797

            Bug ID: 67797
           Summary: [ARM] Unnecessary r0 saving for memset call
           Product: gcc
           Version: 5.2.0
            Status: UNCONFIRMED
          Severity: normal
          Priority: P3
         Component: target
          Assignee: unassigned at gcc dot gnu.org
          Reporter: d.salikhov at samsung dot com
  Target Milestone: ---

Compiled with -Os and with -O2 (result is the same):
#include <string.h>

void *my_func(void *s, size_t n)
{
    memset(s, 0, n);
    return s;
}

The generated code is following:

00000000 <my_func>:
   0:   e92d4010        push    {r4, lr}
   4:   e1a02001        mov     r2, r1
   8:   e1a04000        mov     r4, r0
   c:   e3a01000        mov     r1, #0
  10:   ebfffffe        bl      0 <memset>
  14:   e1a00004        mov     r0, r4
  18:   e8bd8010        pop     {r4, pc}

First, copying r0 into r4 is redundant as memset doesn't clobber r0. So the
code could be reduced to:

push    {r4, lr}
mov     r2, r1
mov     r1, #0
bl      0 <memset>
pop     {r4, pc}

That in turn can be simplified more to:

mov     r2, r1
mov     r1, #0
b       0 <memset>