How does the internal implementation of memcpy work?

Question

How does the standard C function 'memcpy' work? It has to copy a (large) chunk of RAM to another area in the RAM. Since I know you cannot move straight from RAM to RAM in assembly (with the mov instruction) so I am guessing it uses a CPU register as the intermediate memory when copying?

But how does it copy? By blocks (how would it copy by blocks?), by individual bytes (char) or the largest data type they have (copy in long long double's - which is 12 bytes on my system).

EDIT: Ok apparently you can move data from RAM to RAM directly, I am not an assembly expert and all I have learnt about assembly is from this document (X86 assembly guide) which mentions in the section about the mov instruction that you cannot move from RAM to RAM. Apparently this isn't true.

Answer 1

Depends. In general, you couldn't physically copy anything larger than the largest usable register in a single cycle, but that's not really how machines work these days. In practice, you really care less about what the CPU is doing and more about the characteristics of DRAM. The memory hierarchy of the machine is going to play a crucial determining role in performing this copy in the fastest possible manner (e.g., are you loading whole cache-lines? What's the size of a DRAM row with respect to the copy operation?). An implementation might instead choose to use some kind of vector instructions to implement memcpy. Without reference to a specific implementation, it's effectively a byte-for-byte copy with a one-place buffer.

Here's a fun article that describes one person's adventure into optimizing memcpy. The main take-home point is that it is always going to be targeted to a specific architecture and environment based on the instructions you can execute inexpensively.