I couldn't understand if part they do for integers. i < len/sizeof(long). Why is this calculation required ?
Because they are copying words, not individual bytes, in this case (as the comment says, it is an optimization - it requires less iterations and the CPU can handle word aligned data more efficiently).
len is the number of bytes to copy, and sizeof(long) is the size of a single word, so the number of elements to copy (means, loop iterations to execute) is len / sizeof(long).
I couldn't understand if part they do for integers. i < len/sizeof(long). Why is this calculation required ?
Because they are copying words, not individual bytes, in this case (as the comment says, it is an optimization - it requires less iterations and the CPU can handle word aligned data more efficiently).
len is the number of bytes to copy, and sizeof(long) is the size of a single word, so the number of elements to copy (means, loop iterations to execute) is len / sizeof(long).
to understand how it differs from using a loop. But I couldn't any difference of using a loop rather than memcpy, as memcpy uses loop again internally to copy
Well then it uses a loop. Maybe other implementations of libc doesn't do it like that. Anyway, what's the problem/question if it does use a loop? Also as you see it does more than a loop: it checks for alignment and performs a different kind of loop depending on the alignment.
I couldn't understand if part they do for integers. i < len/sizeof(long). Why is this calculation required ?
This is checking for memory word alignment. If the destination and source addresses are word-aligned, and the length copy is multiple of word-size, then it performs an aligned copy by word (long), which is faster than using bytes (char), not only because of the size, but also because most architectures do word-aligned copies much faster.