Why is an empty function not just a return
assembly, compilation, gcc, x86
Solution
Why the rep ?
The reasons are explained in this blog post. In short, jumping directly to a single-byte `ret` instruction would mess up the branch prediction on some AMD processors. And rather than adding a `nop` before the `ret`, a meaningless prefix byte was added to save instruction decoding bandwidth.
The rep is absent in non-empty functions.
To quote from the blog post I linked to: "[`rep ret`] is preferred to the simple `ret` either when it is the target of any kind of branch, conditional (`jne/je/...`) or unconditional (`jmp/call/...`)". In the case of an empty function, the `ret` would have been the direct target of a `call`. In a non-empty function, it wouldn't be.
Why does gcc not remove everything but the ret?
It's possible that some compilers won't omit frame pointer code even if you've specified `-O2`. At least with gcc, you can explicitly tell the compiler to omit them by using the `-fomit-frame-pointer` option.
Problem
If I compile an empty C function ``` void nothing(void) { } ``` using `gcc -O2 -S` (and `clang`) on MacOS, it generates: ``` _nothing: pushq %rbp movq %rsp, %rbp popq %rbp ret ``` Why does `gcc` not remove everything but the `ret`? It seems like an easy optimisation to make unless it really does something (seems not to, to me). This pattern (push/move at the beginning, pop at the end) is also visible in other non-empty functions where `rbp` is otherwise unused. On Linux using a more recent `gcc` (4.4.5) I see just ``` nothing: rep ret ``` Why the `rep` ? The `rep` is absent in non-empty functions.