c - 使用 fork() 的单词频率

Question

我有以下程序，但在某个时候卡住了。(该程序不会创建更多的子进程，而只会创建一个。有人能帮我吗？

Write a program that counts occurrences of a string as a substring in another string (the two strings are given as arguments on the command line). Every time it checks if the first string appears as a substring starting from a position, checking will be done by a child process (obtained with fork) and the father process is not expecting for the child process to finish to initiate a search starting from a different position - so the verifications are being made in parallel. Each child process returns 0 = did not checked (not shown as substring from that position), 1 = has been verified. After conducting all searches, the father process is expected to finish all sons processes and gather their return codes - this value is will be printed(is the number of times as a substring).

#include<stdio.h>
#include<stdlib.h>
#include<unistd.h>
#include<string.h>
char *s1,*s2;
int verificare(char *s1, char *s2, int lungime)
{
  pid_t pid;
  int i,status;
  for (i = 0; i < lungime; i++) {
    pid = fork();
    switch (pid) {
    case -1:
      return EXIT_FAILURE;

    case 0: 
      if (strstr(s1, s2) != NULL)
        return 1;
      else
        return 0;
    _exit(0);

    default:
        waitpid(pid, &status, 0);
    if (WIFEXITED(status))
      printf("Child %d, Return code %d\n", pid,
             WEXITSTATUS(status));
    return status;
}
  }

}
int main(int argc,char **argv){
    if (argc!=3){
        printf("Too less arg");
        return 0;
    }
    s1=argv[1];
    s2=argv[2];
verificare(s1,s2,strlen(s2));
return 0;
}

Answer 1

1. 您正在使用 strstr，它将执行您自己的程序应该执行的操作。您不应该循环遍历以下字符并将它们与模式进行比较吗？
2. 您的程序基本上是按顺序运行的，因为您正在wait()等待子进程在您启动子进程后立即完成。您不应该将所有子 PID 存储在某个数组中，并且只有在所有子 PID 启动后才开始等待吗？
3. 您的父进程永远不会在子进程中看到 verificare 的返回值，因为您的 main 会忽略它并始终返回 0。也许使用 exit 从 child 那里返回？
4. 您的程序无法向用户报告是否找到了匹配项。

我知道这是一项家庭作业，你不应该按照他们要求的方式去做。但是请注意，这不是 fork 的一个很好的用法。可以使用单个处理器检查子字符串的出现，时间与要搜索的字符串的长度成线性关系(使用有限自动机)。所以这个并行算法在这里并没有真正给你买任何东西。