本文介绍了C-迭代多叉的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我需要做四把叉子1000次.我写了这个,但是它永远运行:

I need to make 4 forks 1000 times. I wrote this but it runs forever:

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

#define   N  512

void  chunk0(void);
void  chunk1(void);
void  chunk2(void);
void  chunk3(void);
double get_time(void);

void  main(void)
{
    int i,j,k,iterations=0;
    unsigned int *a=(unsigned int *)malloc(N*N*(sizeof(unsigned int)));
    unsigned int *b=(unsigned int *)malloc(N*N*(sizeof(unsigned int)));
    unsigned int *c=(unsigned int *)malloc(N*N*(sizeof(unsigned int)));
    pid_t  pid;

    for(iterations=0;iterations<1000;iterations++){
        srand ( time(NULL) );
        double start=get_time();

        pid = fork();
        if (pid == 0) {
             chunk0();
        }else {
            pid = fork();
            if (pid == 0){ 
                 chunk1();
            }else {
                pid = fork();
                if (pid == 0){ 
                     chunk2();
                }else {
                    chunk3();
                    wait(NULL);
                    double end=get_time();
                    double diff=end-start;
                    printf("\n Time for run this code is: %lf seconds \n",diff);
                }
            }          
        }
    }       
}

void  chunk0(void)
{
/*  int i,j,k,iterations=0;
    unsigned int *a=(unsigned int *)malloc(N*N*(sizeof(unsigned int)));
    unsigned int *b=(unsigned int *)malloc(N*N*(sizeof(unsigned int)));
    unsigned int *c=(unsigned int *)malloc(N*N*(sizeof(unsigned int)));

    for(iterations=0;iterations<1000;iterations++){
        //printf("iteration #%d: Generating Matrix - ",iterations+1);
        for(i=0;i<N;i++){
            for(j=0;j<N;j++){
                //give int number between 0...1000 to a[i][j] , b[i][j] and reset c[i][j]
                *(a+(i*N+j))=(rand()%1001);
                *(b+(i*N+j))=(rand()%1001);
                *(c+(i*N+j))=0;
            }
        }
        //printf("Multiplying ... \n");
        for(i=0;i<N;i++){
            for(j=0;j<N;j++){
                for(k=0;k<N;k++){
                    *(c+(i*N+j))= *(c+(i*N+j)) + ((*(a+(i*N+k)))*(*(b+(k*N+j))));
                }
            }
        }
    }

    free(a);
    free(b);
    free(c);
*/  
     printf("   *** Child process 0 is done ***\n");
}
void  chunk1(void)
{
     int   i;
     printf("   *** Child process 1 is done ***\n");
}
void  chunk2(void)
{
     int   i;
     printf("   *** Child process 2 is done ***\n");
}
void  chunk3(void)
{
     int   i;
     printf("   *** Child process 3 is done ***\n");
}

double get_time(void){
    struct timeval stime;
    gettimeofday (&stime, (struct timezone*)0);
    return (stime.tv_sec+((double)stime.tv_usec)/1000000);
}

我知道为什么,但是不知道如何解决

I know why, but don't know how to fix it

推荐答案

因为在每个fork()之后,子进程都会从父进程所在的位置继续执行代码.因此,父进程和子进程都继续运行for循环,不仅父进程而且子进程都继续派生.

Because after each fork() the child process continues the code from where the parent process is. Thus both parent and child processes continue running the for loop and not only parent process but also the child processes continue forking.

这篇关于C-迭代多叉的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!

10-28 16:26