Question

我正在尝试使用clock_gettime比较函数调用，但我得到了奇怪的结果。

#include <sys/time.h> 
#include <unistd.h>
#include <stdio.h>
#include <time.h>
#include <stdint.h>
#include <stdlib.h>


int myppid(){
  int myarray[] = {1,2};
  return(myarray[1]);
}

main(int argc, char **argv){

  uint64_t diff,diff1,diff2;
  struct timespec start,end;
  int billion = 1000000000;
  int i;

  /*** for loop to find average overhead ***/

  clock_gettime(CLOCK_MONOTONIC, &start);   //start clock
  for (i; i<1000000; i++);                  //waste time
  clock_gettime(CLOCK_MONOTONIC, &end);     //end clock

  diff = billion * ( end.tv_sec - start.tv_sec) + end.tv_nsec - start.tv_nsec;

  printf("elapsed time of empty for loop: %llu nanoseconds\n", (long long unsigned int) diff);


  /***  myppid to find average overhead of a local function  ****/

  clock_gettime(CLOCK_MONOTONIC, &start);   //start clock
  for (i; i<1000000; i++)                   //waste time
    {myppid();}
  clock_gettime(CLOCK_MONOTONIC, &end);     //end clock

  diff1 = billion * ( end.tv_sec - start.tv_sec) + end.tv_nsec - start.tv_nsec;
  //diff1 = diff1 - diff;
  printf("elapsed time of for loop of myppid: %llu nanoseconds\n", (long long unsigned int) diff1);


  // getppid.c to find average overhead of the system call


 clock_gettime(CLOCK_MONOTONIC, &start);   //start clock
  for (i; i<1000000; i++)                   //waste time
    {getppid();}
  clock_gettime(CLOCK_MONOTONIC, &end);     //end clock

  diff2 = billion * ( end.tv_sec - start.tv_sec) + end.tv_nsec - start.tv_nsec;
  //diff2 = diff2 - diff1;
  printf("elapsed time of for loop of getppid: %llu nanoseconds\n", (long long unsigned int) diff2);

  exit(0);
}

我除了在纳秒内增加时钟数，但我得到：

为空循环的经过时间：421纳秒
for myppid循环的经过时间：160纳秒
getppid循环的经过时间：195纳秒

Answer 1

#include <sys/time.h> 
#include <unistd.h>
#include <stdio.h>
#include <time.h>
#include <stdint.h>
#include <stdlib.h>


int myppid(){
  int myarray[] = {1,2};
  return(myarray[1]);
}

main(int argc, char **argv){

uint64_t diff,diff1,diff2;
struct timespec start,end;
int billion = 1000000000;
int i;
// getppid.c to find average overhead of the system call


clock_gettime(CLOCK_MONOTONIC, &start);   //start clock
 for (i; i<1000000; i++)                   //waste time
   {getppid();}
 clock_gettime(CLOCK_MONOTONIC, &end);     //end clock

 diff2 = billion * ( end.tv_sec - start.tv_sec) + end.tv_nsec - start.tv_nsec;
 //diff2 = diff2 - diff1;
 printf("elapsed time of for loop of getppid: %llu nanoseconds\n", (long long unsigned     int) diff2);

/***  myppid to find average overhead of a local function  ****/

 clock_gettime(CLOCK_MONOTONIC, &start);   //start clock
 for (i; i<1000000; i++)                   //waste time
 {myppid();}
 clock_gettime(CLOCK_MONOTONIC, &end);     //end clock

 diff1 = billion * ( end.tv_sec - start.tv_sec) + end.tv_nsec - start.tv_nsec;
 //diff1 = diff1 - diff;
 printf("elapsed time of for loop of myppid: %llu nanoseconds\n", (long long unsigned int) diff1);

 /*** for loop to find average overhead ***/

 clock_gettime(CLOCK_MONOTONIC, &start);   //start clock
 for (i; i<1000000; i++);                  //waste time
 clock_gettime(CLOCK_MONOTONIC, &end);     //end clock

 diff = billion * ( end.tv_sec - start.tv_sec) + end.tv_nsec - start.tv_nsec;

 printf("elapsed time of empty for loop: %llu nanoseconds\n", (long long unsigned int) diff);

exit(0); 
}

令人惊讶的是，此代码还会为您提供递减的时间值顺序。但如果你访问this link，那就不那么令人惊讶了。它是一种编译器优化技术。

在C中计算函数调用与系统调用的时间

1 个答案: