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Rather than using a C implementation of memset, directly call memset, which 

typically has a much faster optimized implementation.
This commit is contained in:
Wilco Dijkstra 2015-02-27 14:41:46 +00:00
parent 49f476f479
commit ddcf6798d3
2 changed files with 6 additions and 55 deletions
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4
ChangeLog
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@ -1,3 +1,7 @@
2015-02-27 Wilco Dijkstra wdijkstr@arm.com
* string/bzero.c (__bzero): Call memset for performance.
2015-02-27 John David Anglin <dave.anglin@bell.net> 2015-02-27 John David Anglin <dave.anglin@bell.net>
* sysdeps/unix/sysv/linux/hppa/bits/fcntl.h (__O_SYNC): Change * sysdeps/unix/sysv/linux/hppa/bits/fcntl.h (__O_SYNC): Change

57
string/bzero.c
View File

@ -17,66 +17,13 @@
<http://www.gnu.org/licenses/>. */ <http://www.gnu.org/licenses/>. */
#include <string.h> #include <string.h>
#include <memcopy.h>
#undef __bzero #undef __bzero
/* Set N bytes of S to 0. */ /* Set N bytes of S to 0. */
void void
__bzero (s, len) __bzero (void *s, size_t len)
void *s;
size_t len;
{ {
long int dstp = (long int) s; memset (s, '\0', len);
const op_t zero = 0;
if (len >= 8)
{
size_t xlen;
/* There are at least some bytes to zero. No need to test
for LEN == 0 in this alignment loop. */
while (dstp % OPSIZ != 0)
{
((byte *) dstp)[0] = 0;
dstp += 1;
len -= 1;
}
/* Write 8 op_t per iteration until less than 8 op_t remain. */
xlen = len / (OPSIZ * 8);
while (xlen != 0)
{
((op_t *) dstp)[0] = zero;
((op_t *) dstp)[1] = zero;
((op_t *) dstp)[2] = zero;
((op_t *) dstp)[3] = zero;
((op_t *) dstp)[4] = zero;
((op_t *) dstp)[5] = zero;
((op_t *) dstp)[6] = zero;
((op_t *) dstp)[7] = zero;
dstp += 8 * OPSIZ;
xlen -= 1;
}
len %= OPSIZ * 8;
/* Write 1 op_t per iteration until less than op_t remain. */
xlen = len / OPSIZ;
while (xlen != 0)
{
((op_t *) dstp)[0] = zero;
dstp += OPSIZ;
xlen -= 1;
}
len %= OPSIZ;
}
/* Write the last few bytes. */
while (len != 0)
{
((byte *) dstp)[0] = 0;
dstp += 1;
len -= 1;
}
} }
weak_alias (__bzero, bzero) weak_alias (__bzero, bzero)