c - 为什么 GCC 会根据常量的值生成不同的乘法操作码？

Question

我在 gcc.godbolt.org 上玩 GCC 反汇编程序我注意到从 4.6 版开始的 GCC 以不同的方式编译乘法。我有以下两个功能:

unsigned m126(unsigned i)
{
    return i * 126;
}

unsigned m131(unsigned i)
{
    return i * 131;
}

m126 编译成:

mov eax, edi
mov edx, 126
imul eax, edx
ret

而m131编译成:

imul eax, edi, 131
ret

为什么不同？ GCC 4.5 在这两种情况下生成相同的操作码。

A link to the actual example on GCC Explorer .

Answer 1

在 gcc/config/i386/i386.md 中找到这个(见顶部的评论):

;; imul $8/16bit_imm, regmem, reg is vector decoded.
;; Convert it into imul reg, reg
;; It would be better to force assembler to encode instruction using long
;; immediate, but there is apparently no way to do so.
(define_peephole2
  [(parallel [(set (match_operand:SWI248 0 "register_operand")
           (mult:SWI248
            (match_operand:SWI248 1 "nonimmediate_operand")
            (match_operand:SWI248 2 "const_int_operand")))
          (clobber (reg:CC FLAGS_REG))])
   (match_scratch:SWI248 3 "r")]
  "TARGET_SLOW_IMUL_IMM8 && optimize_insn_for_speed_p ()
   && satisfies_constraint_K (operands[2])"
  [(set (match_dup 3) (match_dup 2))
   (parallel [(set (match_dup 0) (mult:SWI248 (match_dup 0) (match_dup 3)))
          (clobber (reg:CC FLAGS_REG))])]
{
  if (!rtx_equal_p (operands[0], operands[1]))
    emit_move_insn (operands[0], operands[1]);
})

好像跟指令解码有关(不好意思我不是专家)