On Wed, 29 Nov 2017, Mason wrote:
Hello,
Consider the following code:
#include <limits.h>
#if CHAR_BIT != 8
#error "Only octets (8-bit bytes) are supported"
#endif
static const char map[16] = "xyz";
void foo(unsigned char *src, char *buf)
{
int hi = src[0] / 16;
int lo = src[0] % 16;
buf[0] = map[hi];
buf[1] = map[lo];
}
$ gcc-7 -O2 -S -march=haswell -Wall -Wextra foo.c
foo:
movzbl (%rdi), %eax # eax = src[0]
movl %eax, %edx # edx = src[0]
andl $15, %eax # eax = src[0] % 16 = lo
shrb $4, %dl # edx = src[0] / 16 = hi
movzbl map(%rax), %eax # eax = map[lo]
andl $15, %edx # Unnecessary, see discussion below
movzbl map(%rdx), %edx
movb %al, 1(%rsi)
movb %dl, (%rsi)
ret
Since CHAR_BIT is 8, it is guaranteed that 0 <= src[0] < 256
therefore 0 <= hi < 16
therefore "andl $15, %edx" is a no-op and could be removed.
In fact, if I change the program to:
#include <limits.h>
#if CHAR_BIT != 8
#error "Only octets (8-bit bytes) are supported"
#endif
static const char map[16] = "xyz";
void foo(unsigned char *src, char *buf)
{
int hi = src[0] / 16;
int lo = src[0] % 16;
buf[0] = hi;
buf[1] = map[lo];
}
Then the generated code is
foo:
movzbl (%rdi), %eax
movl %eax, %edx
andl $15, %eax
movzbl map(%rax), %eax
shrb $4, %dl
movb %dl, (%rsi)
movb %al, 1(%rsi)
ret
In that case, GCC doesn't emit the unnecessary instruction.
Is this easily fixable? (FWIW, trunk on godbolt has the same behavior.)
Writing
long hi = src[0];
hi /= 16;
also helps, the extra &15 comes from extending from unsigned char to long.
I don't know how easy/hard it is to improve, but I am sure it is more
likely to happen if you file an issue in bugzilla ;-)
--
Marc Glisse
