7

3 つの子プロセスを作成することになっているこのコードがあり、それぞれが小さな数学演算を実行します。次に、親はすべての子プロセスの結果を使用して最終的な回答を得るはずですが、親の子から実際に結果を読み取る方法が見つかりません。これを行う方法はありますか？

#include <stdio.h>
#include <unistd.h>
#include <stdlib.h>

int main(void)
{
   int pid1, pid2, pid3, status;
   int a=1, b=2, c=5, d=4, e=6, f=3, g;
   int t1, t2, t3;

   printf("Hello World!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!\n");
   printf("Here I am  before use of forking\n");
   printf("I am the PARENT process and pid is : %d\n",getpid());

   pid1 = fork( );
   if (pid1 == 0)
   {      
    printf("\n\nHere I am just after child forking1\n");
    printf("I am the Child process and pid1 is :%d\n",getpid());      
    printf("My parent's pid is :%d\n",getppid());   
    t1 = a+b;
    printf("The answer for t1 is: %d\n", t1);       
    exit(0);
   }
   else
   {
    wait(&status);
        printf("\nHere I am just after parent forking1\n");
        printf("I am the Parent process and pid is: %d\n",getpid());
   }

   pid2 = fork( );
   if (pid2 == 0)
   {      
    printf("\n\nHere I am just after child forking2\n");
    printf("I am the Child process and pid2 is :%d\n",getpid());      
    printf("My parent's pid is :%d\n",getppid());   
    t2 = c+d;
    printf("The answer for t2 is: %d\n", t2);   
    exit(0);    
   }
   else
   {
    wait(&status);
        printf("\nHere I am just after parent forking2\n");
        printf("I am the Parent process and pid is: %d\n",getpid());
   }

   pid3 = fork( );
   if (pid3 == 0)
   {      
    printf("\n\nHere I am just after child forking3\n");
    printf("I am the Child process and pid3 is :%d\n",getpid());      
    printf("My parent's pid is :%d\n",getppid());   
    t3 = e/f;   
    printf("The answer for t3 is: %d\n", t3);   
    exit(0);
   }
   else
   {
    wait(&status);
        printf("\nHere I am just after parent forkingALL\n");
        printf("I am the Parent process and pid is: %d\n",getpid());
   }


   printf("\n\nThe final answer for t1 is: %d\n", t1);
   printf("The final answer for t2 is: %d\n", t2);
   printf("The final answer for t3 is: %d\n", t3);


   g = t1*t2-t3;
   printf("The final answer for g is: %d\n", g);
}
4

4 に答える 4

7

これは、共有メモリという非常に簡単な手法で行うことができます。それがどのように機能するかの完全な例を挙げます。

まず、フィボナッチ数列の最初の項をn出力するプログラムを書きたいとしましょう(そうするのはそれほど論理的ではないことはわかっていますが、簡単な例なので誰でも理解できます)。

  1. n最初の項を表す整数値を読み取る親があります
  2. 次に、親プロセスが子を作成して渡しnます
  3. 次に、子は最初の n 項を計算し、それらを親に返す必要があります。
#include <stdio.h>
#include <stdlib.h>
#include <sys/mman.h>
#include <fcntl.h>
#include <sys/stat.h>

void printFibo(int n, int *fibo)
{
    int i;
    for(i=0; i<=n; i++)
        printf("%d -> %d\n" ,i, fibo[i]);
}

void computeFibo(int n, int *fibo) 
{
    int i;
    fibo[0] = 0;
    fibo[1] = 1;

    for (i=2; i<=n; i++) 
        fibo[i] = fibo[i-1] + fibo[i-2];
}

int main(int argc, char *argv[])
{
    pid_t childPID;
    int status;
    int shm_fd;
    int* shared_memory;
    int msize; // the size (in bytes) of the shared memory segment 
    const char *name = "FIBONACCI_SERIES";
    int n;

    if (argc!=2) 
    {
        fprintf(stderr, "usage: %s <Fibonacci number to be generated>\n", argv[0]);
        return -1;
    }

    n = atoi(argv[1]);
    if (n < 0) 
    {
        fprintf(stderr, "Illegal fibonacci number: %s\n", argv[1]);
        return -2;
    }

    // calculating the array size based on the number of terms being passed from child to parent
    msize = (n+2)*sizeof(int); 

    // open the memory
    shm_fd = shm_open (name, O_CREAT | O_EXCL | O_RDWR, S_IRWXU | S_IRWXG);
    if (shm_fd < 0) 
    {
        fprintf(stderr,"Error in shm_open()");
        return -3;
    }

    printf("Created shared memory object %s\n", name);

    // attach the shared memory segment
    ftruncate(shm_fd, msize);
    printf("shmat returned\n");

    // allocating the shared memory
    shared_memory = (int *) mmap(NULL, msize, PROT_READ | PROT_WRITE, MAP_SHARED, shm_fd, 0);
    if (shared_memory == NULL) 
    {
        fprintf(stderr,"Error in mmap()");
        return -3;
    }

    printf("Shared memory segment allocated correctly (%d bytes).\n", msize);

    shared_memory[0] = n;

    childPID=fork();
    if ( childPID == -1 ) 
    {
        fprintf(stderr, "Cannot proceed. fork() error");
        return -4;
    }
    if (childPID  == 0) 
    {
        // then we're the child process
        computeFibo(shared_memory[0],shared_memory+1);
        exit(0);
    }
    else
    {
        // parent will wait until the child finished
        wait(&status);

        // print the final results in the 
        printFibo(shared_memory[0], shared_memory+1);

        // now detach the shared memory segment
        shm_unlink(name);
    }
    return 0;
}
于 2014-05-25T22:58:55.003 に答える
2

forkプロセスのコピーを作成するため、子プロセスを呼び出すとfork、親から読み取ることが期待される変数t1、t2、およびt3の独自のコピーがあります。

ですから、あなたがexit子供になると、子供たちは彼らに局所的な計算値とともに死にます。

子から値を読み取りたい場合は、pipesメモリを使用または共有する必要があります。

于 2012-09-19T17:33:14.343 に答える
1

You have to create pipe in the parent process, than after fork you must to close input file descriptor in the child process and close output file descriptor in the parent process.

There is example from the pipe(2) man page.

   #include <sys/wait.h>
   #include <stdio.h>
   #include <stdlib.h>
   #include <unistd.h>
   #include <string.h>

   int
   main(int argc, char *argv[])
   {
       int pipefd[2];
       pid_t cpid;
       char buf;

       if (argc != 2) {
        fprintf(stderr, "Usage: %s <string>\n", argv[0]);
        exit(EXIT_FAILURE);
       }

       if (pipe(pipefd) == -1) {
           perror("pipe");
           exit(EXIT_FAILURE);
       }

       cpid = fork();
       if (cpid == -1) {
           perror("fork");
           exit(EXIT_FAILURE);
       }

       if (cpid == 0) {    /* Child reads from pipe */
           close(pipefd[1]);          /* Close unused write end */

           while (read(pipefd[0], &buf, 1) > 0)
               write(STDOUT_FILENO, &buf, 1);

           write(STDOUT_FILENO, "\n", 1);
           close(pipefd[0]);
           _exit(EXIT_SUCCESS);

       } else {            /* Parent writes argv[1] to pipe */
           close(pipefd[0]);          /* Close unused read end */
           write(pipefd[1], argv[1], strlen(argv[1]));
           close(pipefd[1]);          /* Reader will see EOF */
           wait(NULL);                /* Wait for child */
           exit(EXIT_SUCCESS);
       }
   }
于 2012-09-19T17:42:44.493 に答える