Why does this code generate much more assembly tha

问题:

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Closed 2 years ago.

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?

回答1:

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).

回答2:

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.

回答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)

回答4:

It doesn't (the actual difference is much smaller than shown in the question). I'm surprised nobody checked the C++ output:

godbolt C++ x64 clang 4.0, no compiler options

godbolt Rust 1.18, no compiler options