Why does this code generate much more assembly than equivalent C++/Clang?

Compiling with the compiler flag -O (and with an added pub), I get this output (Link to Godbolt):

push    rbp
mov     rbp, rsp
xor     dil, 1
or      dil, sil
mov     eax, edi
pop     rbp
ret

A few things:

Why is it still longer than the C++ version?

The Rust version is exactly three instructions longer:
```
push    rbp
mov     rbp, rsp
[...]
pop     rbp
```
These are instructions to manage the so called frame pointer or base pointer (rbp). This is mainly required to get nice stack traces. If you disable it for the C++ version via -fno-omit-frame-pointer, you get the same result. Note that this uses g++ instead of clang++ since I haven't found a comparable option for the clang compiler.
Why doesn't Rust omit frame pointer?

Actually, it does. But Godbolt adds an option to the compiler to preserve frame pointer. You can read more about why this is done here. If you compile your code locally with rustc -O --crate-type=lib foo.rs --emit asm -C "llvm-args=-x86-asm-syntax=intel", you get this output:
```
f1:
    xor dil, 1
    or  dil, sil
    mov eax, edi
    ret
```
Which is exactly the output of your C++ version.

You can "undo" what Godbolt does by passing -C debuginfo=0 to the compiler.
Why -O instead of --release?

Godbolt uses rustc directly instead of cargo. The --release flag is a flag for cargo. To enable optimizations on rustc, you need to pass -O or -C opt-level=3 (or any other level between 0 and 3).

To get the same asm code, you need to disable debug info - this will remove the frame pointers pushes.

-C opt-level=3 -C debuginfo=0 (https://godbolt.org/g/vdhB2f)

Compiling with -C opt-level=3 in godbolt gives:

example::f1:
  push rbp
  mov rbp, rsp
  xor dil, 1
  or dil, sil
  mov eax, edi
  pop rbp
  ret

Which looks comparable to the C++ version. See Lukas Kalbertodt's answer for more explanation.

Note: I had to make the function pub extern to stop the compiler optimising it to nothing, as it is unused.

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