关于fill函数和memset函数的效率问题

御笔剑客

在同样的环境下，用fill函数和memset函数填充同一个数组，哪个效率更高呢？ fill函数和for循环比较呢？

行客

memset>for>fill

memset约等于用{0}初始化数组

御笔剑客

行客 发表于 2019-1-28 08:32
memset>for>fill

memset约等于用{0}初始化数组

为神马fill函数的效率会比for循环差呢？

人造人

在同样的环境下，用fill函数和memset函数填充同一个数组，哪个效率更高呢？ fill函数和for循环比较呢？

答案是不确定，不知道
完全取决于这个环境下的fill函数和memset函数的实现

下面贴两个memset函数的实现

1. /* Copyright (C) 1991 Free Software Foundation, Inc.
   
2. This file is part of the GNU C Library.
   
3. 
   
4. The GNU C Library is free software; you can redistribute it and/or
   
5. modify it under the terms of the GNU Library General Public License as
   
6. published by the Free Software Foundation; either version 2 of the
   
7. License, or (at your option) any later version.
   
8. 
   
9. The GNU C Library is distributed in the hope that it will be useful,
   
10. but WITHOUT ANY WARRANTY; without even the implied warranty of
    
11. MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU
    
12. Library General Public License for more details.
    
13. 
    
14. You should have received a copy of the GNU Library General Public
    
15. License along with the GNU C Library; see the file COPYING.LIB.  If
    
16. not, write to the Free Software Foundation, Inc., 675 Mass Ave,
    
17. Cambridge, MA 02139, USA.  */
    
18. 
    
19. #include <ansidecl.h>
    
20. #include <string.h>
    
21. #include <memcopy.h>
    
22. 
    
23. PTR
    
24. DEFUN(memset, (dstpp, c, len), PTR dstpp AND int c AND size_t len)
    
25. {
    
26.   long int dstp = (long int) dstpp;
    
27. 
    
28.   if (len >= 8)
    
29.     {
    
30.       size_t xlen;
    
31.       op_t cccc;
    
32. 
    
33.       cccc = (unsigned char) c;
    
34.       cccc |= cccc << 8;
    
35.       cccc |= cccc << 16;
    
36.       if (OPSIZ > 4)
    
37.         cccc |= cccc << 32;
    
38. 
    
39.       /* There are at least some bytes to set.
    
40.          No need to test for LEN == 0 in this alignment loop.  */
    
41.       while (dstp % OPSIZ != 0)
    
42.         {
    
43.           ((byte *) dstp)[0] = c;
    
44.           dstp += 1;
    
45.           len -= 1;
    
46.         }
    
47. 
    
48.       /* Write 8 `op_t' per iteration until less than 8 `op_t' remain.  */
    
49.       xlen = len / (OPSIZ * 8);
    
50.       while (xlen > 0)
    
51.         {
    
52.           ((op_t *) dstp)[0] = cccc;
    
53.           ((op_t *) dstp)[1] = cccc;
    
54.           ((op_t *) dstp)[2] = cccc;
    
55.           ((op_t *) dstp)[3] = cccc;
    
56.           ((op_t *) dstp)[4] = cccc;
    
57.           ((op_t *) dstp)[5] = cccc;
    
58.           ((op_t *) dstp)[6] = cccc;
    
59.           ((op_t *) dstp)[7] = cccc;
    
60.           dstp += 8 * OPSIZ;
    
61.           xlen -= 1;
    
62.         }
    
63.       len %= OPSIZ * 8;
    
64. 
    
65.       /* Write 1 `op_t' per iteration until less than OPSIZ bytes remain.  */
    
66.       xlen = len / OPSIZ;
    
67.       while (xlen > 0)
    
68.         {
    
69.           ((op_t *) dstp)[0] = cccc;
    
70.           dstp += OPSIZ;
    
71.           xlen -= 1;
    
72.         }
    
73.       len %= OPSIZ;
    
74.     }
    
75. 
    
76.   /* Write the last few bytes.  */
    
77.   while (len > 0)
    
78.     {
    
79.       ((byte *) dstp)[0] = c;
    
80.       dstp += 1;
    
81.       len -= 1;
    
82.     }
    
83. 
    
84.   return dstpp;
    
85. }
    

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1. static inline void * memset(void * s,char c,int count)
   
2. {
   
3. __asm__("cld\n\t"
   
4.         "rep\n\t"
   
5.         "stosb"
   
6.         ::"a" (c),"D" (s),"c" (count)
   
7.         );
   
8. return s;
   
9. }
   

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