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Home/ Questions/Q 8672571
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Editorial Team
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Editorial Team
Asked: June 12, 20262026-06-12T19:14:19+00:00 2026-06-12T19:14:19+00:00

hey there i’m trying to calculate execution time of a simple multithreaded programme on

  • 0

hey there i’m trying to calculate execution time of a simple multithreaded programme on UBUNTU. despite of i researched and used various methods, i always get 0.0000 value. here is my simple code

#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 );

}

i will appreciate if you can help and thanks anyway.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-12T19:14:20+00:00Added an answer on June 12, 2026 at 7:14 pm

    It looks like you want to time a function in code. Consider 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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