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嘿，我正在尝试计算一个简单的多线程程序在 UBUNTU 上的执行时间。尽管我研究并使用了各种方法，但我总是得到 0.0000 的值。这是我的简单代码

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

clock_t startm, stopm;
#define START if ( (startm = clock()) == -1) {printf("Error calling clock");exit(1);}
#define STOP if ( (stopm = clock()) == -1) {printf("Error calling clock");exit(1);}
#define PRINTTIME printf( "%6.9f seconds used by the processor.\n", ((double)stopm-startm)/CLOCKS_PER_SEC);

void* thread_function(int);
void function();

int total=0;
int counter;

pthread_mutex_t mutex1=PTHREAD_MUTEX_INITIALIZER;

int main(int argc, char* argv[]) {

    START
    counter = atoi(argv[1]);
    function();
    STOP
    PRINTTIME
    return;
}

void function(){
    int i;

    pthread_t t_array[counter];

    for(i=0; i<counter; i++){
        pthread_create(&t_array[i], NULL, thread_function, i);
    }

    for(i=0; i<counter; i++){
        pthread_join(t_array[i],NULL);
    }
    printf("Total = %d\n", total);
}

void* thread_function(int index){

   pthread_mutex_lock( &mutex1 );

   printf("Index : %d\n", index);
   total++;
   usleep(100000);

   pthread_mutex_unlock( &mutex1 );

}

如果您能提供帮助，我将不胜感激，无论如何，谢谢。

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1 回答 1

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看起来您想在代码中为函数计时。考虑 gettimeofday()

#include <sys/types.h>
#include <sys/time.h>
#include <stdlib.h>
#include <stdio.h>

double now(void)
{
   struct timeval tv;
   double retval=0;
   gettimeofday(&tv, NULL);
   retval=tv.tv_usec;
   retval+= (double)tv.tv_usecs / 1000000.;
   return retval;
}

int main()
{
   double start=now();
   // run code here
   printf("elapsed time = %.6f"\n", now() - start);
   return 0;

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于 2012-10-12T00:13:06.223 回答