c++ - 为什么通过指针进行 int 加法比通过指针进行 int 乘法少一条 x86 指令？

Question

我有以下 C/C++ 代码( compiler explorer link ):

void update_mul(int *x, int *amount) { 
    *x *= *amount; 
}

void update_add(int *x, int *amount) { 
    *x += *amount; 
}

在 clang 和 gcc 下编译为 C 或 C++ 至少 -O1启用，以上转换为这个程序集:

update_mul:                             # @update_mul
        mov     eax, dword ptr [rdi]
        imul    eax, dword ptr [rsi]
        mov     dword ptr [rdi], eax
        ret
update_add:                             # @update_add
        mov     eax, dword ptr [rsi]
        add     dword ptr [rdi], eax
        ret

似乎添加它正在做类似的事情:

register = *amount;
*x += register;

但是对于乘法，它正在做:

register = *x;
register *= *amount;
*x = register;

为什么乘法需要额外的加法指令，或者不需要但只是更快？

Answer 1

IA-32 architecture specification ( alternative single-page link ) 表明对于目标(第一个参数)是内存操作数的 IMUL 根本没有编码:

Encoding               | Meaning
IMUL r/m8*             | AX ← AL ∗ r/m byte.
IMUL r/m16             | DX:AX ← AX ∗ r/m word.
IMUL r/m32             | EDX:EAX ← EAX ∗ r/m32.
IMUL r/m64             | RDX:RAX ← RAX ∗ r/m64.
IMUL r16, r/m16        | word register ← word register ∗ r/m16.
IMUL r32, r/m32        | doubleword register ← doubleword register ∗ r/m32.
IMUL r64, r/m64        | Quadword register ← Quadword register ∗ r/m64.
IMUL r16, r/m16, imm8  | word register ← r/m16 ∗ sign-extended immediate byte.
IMUL r32, r/m32, imm8  | doubleword register ← r/m32 ∗ sign- extended immediate byte.
IMUL r64, r/m64, imm8  | Quadword register ← r/m64 ∗ sign-extended immediate byte.
IMUL r16, r/m16, imm16 | word register ← r/m16 ∗ immediate word.
IMUL r32, r/m32, imm32 | doubleword register ← r/m32 ∗ immediate doubleword.
IMUL r64, r/m64, imm32 | Quadword register ← r/m64 ∗ immediate doubleword.