为什么'mov%eax,%eax;不比不快?

Apparently ,现代处理器可以判断你是否做了一些愚蠢的事情,比如把寄存器移到自己的位置( mov %eax, %eax )并对其进行优化。为了验证这一说法,我运行了以下程序:

#include <stdio.h>
#include <time.h>

static inline void f1() {
   for (int i = 0; i < 100000000; i++)
      __asm__(
            "mov %eax, %eax;"
            "nop;"
            );
}

static inline void f2() {
   for (int i = 0; i < 100000000; i++)
      __asm__(
            "nop;"
            );
}

static inline void f3() {
   for (int i = 0; i < 100000000; i++)
      __asm__(
            "mov %ebx, %eax;"
            "nop;"
            );
}

int main() {
   int NRUNS = 10;
   clock_t t, t1, t2, t3;

   t1 = t2 = t3 = 0;
   for (int run = 0; run < NRUNS; run++) {
      t = clock(); f1(); t1 += clock()-t;
      t = clock(); f2(); t2 += clock()-t;
      t = clock(); f3(); t3 += clock()-t;
   }

   printf("f1() took %f cycles on avg\n", (float) t1/ (float) NRUNS);
   printf("f2() took %f cycles on avg\n", (float) t2/ (float) NRUNS);
   printf("f3() took %f cycles on avg\n", (float) t3/ (float) NRUNS);

   return 0;
}

这给了我:

f1() took 175587.093750 cycles on avg
f2() took 188313.906250 cycles on avg
f3() took 194654.296875 cycles on avg

正如人们所料, f3() 出来最慢。但令人惊讶的是(至少对我来说), f1() 比 f2() .为什么?

使现代化 :使用编译 -falign-loops 在质量上给出了相同的结果:

f1() took 164271.000000 cycles on avg
f2() took 173783.296875 cycles on avg
f3() took 177765.203125 cycles on avg