Hello,
Here's a simple memory copy macro:
#define MYMEMCOPY(dp, sp, len) \
do { \
long __len = len; \
while (--__len >= 0) \
(dp)[__len] = (sp)[__len]; \
} while (0)
void foo(unsigned char *dp, const unsigned char *sp, unsigned long size) {
MYMEMCOPY(dp, sp, size);
}
void bar(unsigned char *dp, const unsigned char *sp) {
MYMEMCOPY(dp, sp, 128);
}
The code fragments generated for the foo and bar functions with -O and -O2
optimizations respectively is as follows:
/* ===== With -O switch ===== */
/* function foo */
.L4:
movzbl -1(%rcx), %eax
movb %al, -1(%rdx)
subq $1, %rcx
subq $1, %rdx
subq $1, %r8
jns .L4
/* function bar */
movl $126, %edx
.L8:
.LBB3:
.loc 1 13 0
movzbl 1(%rdx,%rsi), %eax
movb %al, 1(%rdx,%rdi)
subq $1, %rdx
cmpq $-2, %rdx
jne .L8
/* ===== With -O2 switch =====*/
/* function foo */
.L4:
movzbl -1(%rsi), %eax
addq $1, %rdi
subq $1, %rsi
movb %al, -1(%rcx)
subq $1, %rcx
cmpq %rdx, %rdi
jne .L4
/* function bar */
movl $126, %edx
.L9:
.LBB3:
.loc 1 13 0
movzbl 1(%rdx,%rsi), %eax
movb %al, 1(%rdx,%rdi)
subq $1, %rdx
cmpq $-2, %rdx
jne .L9
Now my questions are:
(i) Why does the compiler generate an addq, cmpq and jne for the foo function
with -O2? Isn't subq/jns more efficient, as seen from the output from -O?
(ii) For function bar, why is the "cmpq $-2, %rdx" instruction generated? Won't
it be better to count down from 128 to 0 instead of 126 to -2?
Here's my OS and compiler version (I'm running a 64-bit FreeBSD):
$ uname -a
FreeBSD xxx 7.0-RELEASE FreeBSD 7.0-RELEASE #0: Wed Nov 12 18:54:21 PST 2008
r...@wc7:/usr/obj/usr/src/sys/SMKERNEL amd64
$ cc --version
cc (GCC) 4.2.1 20070719 [FreeBSD]
Copyright (C) 2007 Free Software Foundation, Inc.
This is free software; see the source for copying conditions. There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
And these are the commands I used to compile the program:
cc -S -O -g test.c
cc -S -O2 -g test.c
Any pointers would be appreciated. Thanks!
Regards,
N Datta
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