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

I wrote a simple C++ function in order to check compiler optimization:

bool f1(bool a, bool b) {
    return !a || (a && b);
}

After that I checked the equivalent in Rust:

fn f1(a: bool, b: bool) -> bool {
    !a || (a && b)
}

I used godbolt to check the assembler output.

The result of the C++ code (compiled by clang with -O3 flag) is following:

f1(bool, bool):                                # @f1(bool, bool)
    xor     dil, 1
    or      dil, sil
    mov     eax, edi
    ret

And the result of Rust equivalent is much longer:

example::f1:
  push rbp
  mov rbp, rsp
  mov al, sil
  mov cl, dil
  mov dl, cl
  xor dl, -1
  test dl, 1
  mov byte ptr [rbp - 3], al
  mov byte ptr [rbp - 4], cl
  jne .LBB0_1
  jmp .LBB0_3
.LBB0_1:
  mov byte ptr [rbp - 2], 1
  jmp .LBB0_4
.LBB0_2:
  mov byte ptr [rbp - 2], 0
  jmp .LBB0_4
.LBB0_3:
  mov al, byte ptr [rbp - 4]
  test al, 1
  jne .LBB0_7
  jmp .LBB0_6
.LBB0_4:
  mov al, byte ptr [rbp - 2]
  and al, 1
  movzx eax, al
  pop rbp
  ret
.LBB0_5:
  mov byte ptr [rbp - 1], 1
  jmp .LBB0_8
.LBB0_6:
  mov byte ptr [rbp - 1], 0
  jmp .LBB0_8
.LBB0_7:
  mov al, byte ptr [rbp - 3]
  test al, 1
  jne .LBB0_5
  jmp .LBB0_6
.LBB0_8:
  test byte ptr [rbp - 1], 1
  jne .LBB0_1
  jmp .LBB0_2

I also tried with -O option but the output is empty (deleted unused function).

I intentionally am NOT using any library in order to keep output clean. Please notice that both clang and rustc use LLVM as a backend. What explains this huge output difference? And if it is only disabled-optimize-switch problem, how can I see optimized output from rustc?

Solution

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).