memcpy和memmove的区别

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 1 ＃i nclude 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; }