Linux- 如何算出函数的运行时间_linux计算函数执行时间-优快云博客

背景知识：

1. time() 函数 - get time in seconds

Synopsis

#include <time.h>

time_t time(time_t *t);

Description

time () returns the time as the number of seconds since the Epoch, 1970-01-01 00:00:00 +0000 (UTC).

If t is non-NULL, the return value is also stored in the memory pointed to by t.

Return value:

On success, the value of time in seconds since the Epoch is returned. On error, ((time_t) -1) is returned, and errno is set appropriately.

2. clock() 函数：

NAME

       clock - determine processor time

SYNOPSIS

       #include <time.h>

       clock_t clock(void);

DESCRIPTION

       The clock() function returns an approximation of processor time used
       by the program.

RETURN VALUE

       The value returned is the CPU time used so far as a clock_t; to get
       the number of seconds used, divide by CLOCKS_PER_SEC.  If the
       processor time used is not available or its value cannot be
       represented, the function returns the value (clock_t) -1.

网上有人用time() 函数，想得到Fibo(10)的运行时间：

#include <sys/time.h>
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
void main(int argc,char* argv[]){
  int n;
  if(argc == 2){
    n = atoi(argv[1]);
  }
  struct timeval start, end;
  gettimeofday(&start, 0);
  int r = fib(n);
  gettimeofday(&end, 0);
  long mtime, s,us;
  s = end.tv_sec  - start.tv_sec;
  us = end.tv_usec - start.tv_usec;
  printf("s=%f,us=%f  \n", s, us);
  mtime = (s*1000 + us/1000.0)+0.5;
  printf("Fib result for %d is: %d;elapsing %f \n", n, r, mtime); 

}

int fib(int n){
  if(n == 0) return 0;
  if(n == 1) return 1;
  return fib(n-1)+fib(n-2);
}

问题是：由于函数运行的时间可能非常短，而time() 是以秒为单位的，所以经常得到结果是0.

所以必须多次运行函数，然后得到函数运行时间的平均值：

下面是用clock() 来得到函数运行的平均时间：

#include <stdio.h>
#include <unistd.h>
#include <time.h>
long fib(int n){
  if(n == 0) return 0;
  if(n == 1) return 1;
  return fib(n-1)+fib(n-2);
}

int main ()
{
  int i=0;
  int p = (int) getpid();
  clock_t cstart = clock();
  clock_t cend = 0;
  for (i=0; i<1000000; i++) {
    long f = fib(i%16);
    if (i % p == 0) printf("i=%d, f=%ld\n", i, f);
  }
  cend = clock();
  printf ("%.3f cpu sec\n", ((double)cend - (double)cstart)* 1.0e-6);   //类似于：((double)cend - (double)cstart)/CLOCKS_PER_SEC
  return 0;
}