FMAF怪异的表现

我使用fmaf功能在*和+的使用情况下经历了巨大的性能降低。我在两台Linux机器上，使用G 4.4.3和G 4.6.3

在两台不同的机器上，如果填充myOut向量而没有使用fmaf。

带有G 4.6.3的服务器和Intel（R）Xeon（R）CPU E5-2650 @ 2.00GHz

$ ./a.out fmaf
Time: 1.55008 seconds.
$ ./a.out muladd
Time: 0.403018 seconds.

带有G 4.4.3的服务器和Intel（R）Xeon（R）CPU X5650 @ 2.67GHz

$ ./a.out fmaf
Time: 0.547544 seconds.
$ ./a.out muladd
Time: 0.34955 seconds.

不应该使用fmaf版本（以避免额外的综述，然后更精确）吗？

#include <stddef.h>
#include <iostream>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <sys/time.h>
int main(int argc, char** argv) {
  if (argc != 2) {
    std::cout << "missing parameter: 'muladd' or 'fmaf'"
              << std::endl;
    exit(-1);
  }
  struct timeval start,stop,result;
  const size_t mySize = 1e6*100;
  float* myA = new float[mySize];
  float* myB = new float[mySize];
  float* myC = new float[mySize];
  float* myOut = new float[mySize];
  gettimeofday(&start,NULL);
  if (!strcmp(argv[1], "muladd")) {
    for (size_t i = 0; i < mySize; ++i) {
      myOut[i] = myA[i]*myB[i]+myC[i];
    }
  } else if (!strcmp(argv[1], "fmaf")) {
    for (size_t i = 0; i < mySize; ++i) {
      myOut[i] = fmaf(myA[i], myB[i], myC[i]);
    }
  } else {
    std::cout << "specify 'muladd' or 'fmaf'" << std::endl;
    exit(-1);
  }
  gettimeofday(&stop,NULL);
  timersub(&stop,&start,&result);
  std::cout << "Time: " <<  result.tv_sec + result.tv_usec/1000.0/1000.0
            << " seconds." << std::endl;
  delete []myA;
  delete []myB;
  delete []myC;
  delete []myOut;
}