The C abstract machine presents i and f as having separate locations in memory, but a compiler that knows that memcopy does can avoid actually saving the target machine registers to the stack, then avoid even copying values between registers.

Thanks. Does that really happen?

Yes.

The compiler is not beholden to the standard library but the standard. So all modern compilers come with a bit of knowledge of how standard library functions like memcpy are supposed to behave and as long as the visible effect is the same it's allowed to do anything it wants. So instead of calling the function memcpy for a size of 4, it can e.g. just use a mov instruction to move the value from a float to an integer register. [1] It does additional analysis on how the values are used and maybe it doesn't even need to move the value at all, but that's the gist of it.

[1] https://gcc.godbolt.org/z/x4P7jE

A good example of this is the string functions in the C stdlib. I’ve reverse engineered some programs where the compiler used x86’s built-in “string” instructions instead of calling out to, say, `strlen`.