4 ms·
That's the correct steps, but that doesn't add anything that the original C code doesn't do. On x86, that could be: mov [ebx], eax add [ecx], eax Whic
by freeone3000 8y ago
That's the correct steps, but that doesn't add anything that the original C code doesn't do.
On x86, that could be:
mov [ebx], eax
add [ecx], eax
Which is an entirely separate set of operands, instructions, and addressing modes. It's still the same steps, but it's not really telling you anything.