爲什麼我的進程不能同時運行？

Question

我的問題是：

1.）我怎樣才能讓父進程總是最後死去？我知道這是沒有完成的，因爲父母是第一個運行的pid，但我不知道如何改變它。

2.）我如何讓我的子進程與他同時執行？我甚至把這個數字看得很高，看看它是不是巧合，但它似乎不是。

編輯：SOLUTIONS

1）加入等待（NULL）在兩次內默認
2.）發生了什麼。用睡眠（1）來證明它。

我的代碼如下

#include <stdio.h> 
int main() { 
    int pid, i; 
    pid = fork(); 
    switch(pid) { 
     case -1: 
      // error 
      printf("Fork error"); 
      break; 
     case 0: 
      // child process 
      printf("First child is born, my pid is %d\n", getpid()); 
      for(i=1; i<10; i++) 
       printf("First child executes iteration %d\n", i); 
      printf("First child dies quietly.\n"); 
      break; 
     default: 
      // parent process 
      printf("Parent process is born, my pid is %d\n", getpid()); 
      pid = fork(); 
      switch(pid) { 
       case -1: 
        // error 
        printf("Fork error"); 
        break; 
       case 0: 
        // child process 
        printf("Second child is born, my pid is %d\n", getpid()); 
        for(i=1; i<10; i++) 
         printf("Second child executes iteration %d\n", i); 
        printf("Second child dies quietly.\n"); 
        break; 
       default: 
        // parent process 
        printf("Parent process dies quietly."); 
     } 
    } 
    return 0; 
} 

我總是輸出看起來像這樣：

 
Parent process is born, my pid is 7847 
First child is born, my pid is 7848 
First child executes iteration: 1 
First child executes iteration: 2 
First child executes iteration: 3 
First child executes iteration: 4 
First child executes iteration: 5 
First child executes iteration: 6 
First child executes iteration: 7 
First child executes iteration: 8 
First child executes iteration: 9 
First child executes iteration: 10 
First child dies quietly. 
Parent process dies quietly. 
Second child is born, my pid is 7849 
Second child executes iteration 1 
Second child executes iteration 2 
Second child executes iteration 3 
Second child executes iteration 4 
Second child executes iteration 5 
Second child executes iteration 6 
Second child executes iteration 7 
Second child executes iteration 8 
Second child executes iteration 9 
Second child executes iteration 10 
Second child dies quietly. 

我的任務是：

編寫一個C語言程序（ 「procs.c」），其創建三個進程：創建兩個子進程的父進程。

的第一個孩子應該做到以下幾點：

• 顯示「第一個孩子出生後，我的pid是......」

• 顯示十倍的消息「第一個孩子來執行迭代X」 ，其中X是迭代次數

• 顯示「第一個孩子安靜地死亡。」

第二個孩子應該做到以下幾點：

• 顯示「第二個孩子出生後，我的pid是......」

• 顯示十倍的消息「老二執行迭代X「，其中X是迭代次數

• 顯示」第二個孩子安靜地死亡。「

父進程應該做到以下幾點：

• 顯示 「父進程是天生的，我的pid是......」

• 創建的第一個孩子

• 創建第二個孩子

• display「Parent過程悄然死亡。「

使用gcc編譯程序並將其命名爲可執行的‘特效’執行程序數次，並注意如何將兩個孩子的輸出隔行

這項計劃的一個可能的輸出是：。

 
nova> ./procs 

Parent process is born, my pid is 7847 
First child is born, my pid is 7848 
First child executes iteration: 1 
First child executes iteration: 2 
First child executes iteration: 3 
First child executes iteration: 4 
First child executes iteration: 5 
Second child is born, my pid is 7849 
Second child executes iteration 1 
Second child executes iteration 2 
Second child executes iteration 3 
First child executes iteration: 6 
Second child executes iteration 4 
Second child executes iteration 5 
Second child executes iteration 6 
First child executes iteration: 7 
Second child executes iteration 7 
Second child executes iteration 8 
Second child executes iteration 9 
Second child executes iteration 10 
Second child dies quietly. 
First child executes iteration: 8 
First child executes iteration: 9 
First child executes iteration: 10 
First child dies quietly. 
Parent process dies quietly. 

Answer 1

1. 合適的父進程應該等待（使用wait() or waitpid()或特定於平臺的變體）其孩子離開之前死亡。

2. 併發執行需要調度程序有機會運行。您可以通過適當的睡眠（微睡眠，nano-sleep）操作來強制解決問題。一些系統調用會有幫助（所以fflush(0)可能會有所幫助）。

---

#include <stdio.h> 
#include <sys/wait.h> 
#include <time.h> 
#include <unistd.h> 

int main(void) 
{ 
    int pid, i; 
    struct timespec tw = { .tv_sec = 0, .tv_nsec = 10000000 }; 
    pid = fork(); 
    switch(pid) 
    { 
     case -1: 
      printf("Fork error"); 
      break; 
     case 0: 
      printf("First child is born, my pid is %d\n", getpid()); 
      for(i=1; i<10; i++) 
      { 
       printf("First child executes iteration %d\n", i); 
       nanosleep(&tw, 0); 
      } 
      printf("First child dies quietly.\n"); 
      break; 
     default: 
      printf("Parent process is born, my pid is %d\n", getpid()); 
      pid = fork(); 
      switch(pid) 
      { 
       case -1: 
        printf("Fork error"); 
        break; 
       case 0: 
        printf("Second child is born, my pid is %d\n", getpid()); 
        for(i=1; i<10; i++) 
        { 
         printf("Second child executes iteration %d\n", i); 
         nanosleep(&tw, 0); 
        } 
        printf("Second child dies quietly.\n"); 
        break; 
       default: 
        printf("Parent process waiting for children.\n"); 
        int corpse; 
        int status; 
        while ((corpse = waitpid(0, &status, 0)) > 0) 
         printf("Child %d died with exit status 0x%.4X\n", corpse, status); 
        printf("Parent process dies quietly.\n"); 
        break; 
      } 
    } 
    return 0; 
} 

輸出示例：

Parent process is born, my pid is 46624 
First child is born, my pid is 46625 
First child executes iteration 1 
Parent process waiting for children. 
Second child is born, my pid is 46626 
Second child executes iteration 1 
First child executes iteration 2 
Second child executes iteration 2 
First child executes iteration 3 
Second child executes iteration 3 
First child executes iteration 4 
Second child executes iteration 4 
Second child executes iteration 5 
First child executes iteration 5 
Second child executes iteration 6 
First child executes iteration 6 
Second child executes iteration 7 
First child executes iteration 7 
Second child executes iteration 8 
First child executes iteration 8 
Second child executes iteration 9 
First child executes iteration 9 
First child dies quietly. 
Second child dies quietly. 
Child 46625 died with exit status 0x0000 
Child 46626 died with exit status 0x0000 
Parent process dies quietly. 

注意，10點毫秒的延遲幾乎迫使交替執行。如果沒有類似的東西，你會被系統和調度程序的特質所困擾，許多現代化的機器太快了！

Answer 2

fork()創建一個新的子進程，從父，其運行的獨立。

如果你想父進程等待孩子們完成後，您可以使用系統調用wait，傳遞一個PID，這將阻塞，直到過程結束。

看文章wait(System Call) on Wikipedia