c - 了解 memcpy() 的源代码

Question

00018 void *memcpy(void *dst, const void *src, size_t len)
00019 {
00020         size_t i;
00021 
00022         /*
00023          * memcpy does not support overlapping buffers, so always do it
00024          * forwards. (Don't change this without adjusting memmove.)
00025          *
00026          * For speedy copying, optimize the common case where both pointers
00027          * and the length are word-aligned, and copy word-at-a-time instead
00028          * of byte-at-a-time. Otherwise, copy by bytes.
00029          *
00030          * The alignment logic below should be portable. We rely on
00031          * the compiler to be reasonably intelligent about optimizing
00032          * the divides and modulos out. Fortunately, it is.
00033          */
00034 
00035         if ((uintptr_t)dst % sizeof(long) == 0 &&
00036             (uintptr_t)src % sizeof(long) == 0 &&
00037             len % sizeof(long) == 0) {
00038 
00039                 long *d = dst;
00040                 const long *s = src;
00041 
00042                 for (i=0; i<len/sizeof(long); i++) {
00043                         d[i] = s[i];
00044                 }
00045         }
00046         else {
00047                 char *d = dst;
00048                 const char *s = src;
00049 
00050                 for (i=0; i<len; i++) {
00051                         d[i] = s[i];
00052                 }
00053         }
00054 
00055         return dst;
00056 }

我刚刚经历了 memcpy 的实现, 了解它与使用循环有何不同。但是我看不出使用循环而不是 memcpy 有什么区别。 , 作为 memcpy在内部再次使用循环进行复制。

我听不懂if他们为整数所做的部分 — i < len/sizeof(long) .为什么需要这个计算？

Answer 1

I couldn't understand if part they do for integers. i < len/sizeof(long). Why is this calculation required ?

因为在这种情况下它们复制的是单词，而不是单个字节(正如评论所说，这是一种优化 - 它需要更少的迭代，CPU 可以更有效地处理单词对齐的数据)。

len 是要复制的字节数，sizeof(long) 是单个单词的大小，因此要复制的元素数(即要执行的循环迭代)为 len/sizeof(long)。