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>
