Understanding the source code of memcpy()

Question

00018 void *memcpy(void *dst, const void *src, size_t len)
00019 {
00020         size_t i;
00021 
00022         /*
00023          * memcpy does not support overlappi         


        

Answer 1

len%sizeof(long) checks if you are trying to copy full-longs not a part of long.

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

checks for alignment and if true, copies fast(sizeof(long) bytes at a time).

00046    else {
00047                 char *d = dst;
00048                 const char *s = src;
00049 
00050                 for (i=0; i

this is for the mis-aligned arrays (slow copy (1 byte at a time))