听风竹轩的书架

区别:
从DESCRIPTION看来，两者的功能基本相同，唯一不同的是，当 dest 和 src 有重叠的时候选用不同的函数可能会造成不同的结果。不妨写个小程序来测一下：
0 ＃i nclude <string.h>
1 ＃i nclude <stdio.h>
2
3 int main()
4 {
5    int i = 0;
6    int a[10];
7
8    for(i; i < 10; i++)
9    {
10        a[i] = i;
11   }
12
13   memcpy(&a[4], a, sizeof(int)*6);
14
15   for(i = 0; i < 10; i++)
16   {
17       printf("%d ",a[i]);
18   }
20
21    printf("\n");
22    return 0;
23 }
很简单的小程序！不过已经足以达到我的目的了：）将上面代码gcc之后再运行，结果为：0 1 2 3 0 1 2 3 0 1 。
再把第13行改成：memmove(&a[4], a, sizeof(int)*6)，重新gcc再运行，结果为：0 1 2 3 0 1 2 3 4 5 ！
呵呵，两者的区别出现了。不过其实这样还不够，继续修改13行： memmove(a, &a[4], sizeof(int)*6) //也就是将源、目的置换一下而已
重新gcc编译再运行，结果为：4 5 6 7 8 9 6 7 8 9 。
还不够，继续修改13行为： memcpy(a, &a[4], sizeof(int)*6); gcc并运行，结果仍为： 4 5 6 7 8 9 6 7 8 9 ！
至此真相已经大白了。对比上面四个结果，不难得出以下结论：
1. 当 src 和 dest 所指内存区有重叠时，memmove 相对 memcpy 能提供保证：保证能将 src 所指内存区的前 n 个字节正确的拷贝到 dest 所指内存中；
2. 当 src 地址比 dest 地址低时，两者结果一样。换句话说，memmove 与 memcpy 的区别仅仅体现在 dest 的头部和 src 的尾部有重叠的情况下；
memcpy

代码:
;***
;memcpy.asm - contains memcpy and memmove routines
;
;       Copyright (c) 1986-1997, Microsoft Corporation. All right reserved.
;
;Purpose:
;       memcpy() copies a source memory buffer to a destination buffer.
;       Overlapping buffers are not treated specially, so propogation may occur.
;       memmove() copies a source memory buffer to a destination buffer.
;       Overlapping buffers are treated specially, to avoid propogation.
;
;*******************************************************************************
;***
;memcpy - Copy source buffer to destination buffer
;
;Purpose:
;       memcpy() copies a source memory buffer to a destination memory buffer.
;       This routine does NOT recognize overlapping buffers, and thus can lead
;       to propogation.
;       For cases where propogation must be avoided, memmove() must be used.
;
;       Algorithm:

       void* memcpy(void* dest, void* source, size_t count)

      {

           void* ret = dest;

          //copy from lower address to higher address

          while (count--)

                  *dest++ = *source;

           return ret;

      }

memmove

memmove - Copy source buffer to destination buffer
;
;Purpose:
;       memmove() copies a source memory buffer to a destination memory buffer.
;       This routine recognize overlapping buffers to avoid propogation.
;       For cases where propogation is not a problem, memcpy() can be used.
;
;   Algorithm:

    void* memmove(void* dest, void* source, size_t count)

   {

       void* ret = dest;

      if (dest <= source || dest >= (source + count))

       {

          //Non-Overlapping Buffers
         //copy from lower addresses to higher addresses
    

         while (count --)

               *dest++ = *source++;

     }

     else

     {

        //Overlapping Buffers
       //copy from higher addresses to lower addresses

       dest += count - 1;

       source += count - 1;

       while (count--)

                *dest-- = *source--;l

     }

      return ret;

   }

另一种实现：

void* mymemcpy( void* dest, const void* src, size_t count )
{
     char* d = (char*)dest;
     const char* s = (const char*)src;
   //   int n = (count + 7) / 8; // count > 0 assumed
     int n = count >> 3;
     switch( count & 7 )
     {
               do {   *d++ = *s++;
     case 7:         *d++ = *s++;
     case 6:         *d++ = *s++;
     case 5:         *d++ = *s++;
     case 4:         *d++ = *s++;
     case 3:         *d++ = *s++;
     case 2:         *d++ = *s++;
     case 1:         *d++ = *s++;
     case 0           } //while (--n > 0);
                  while (n-- > 0)
     }

     return dest;
}