C中的多个1对1管道

Question

C 中是否可以有多个并发的一对一管道？ 我的示例用例如下：

1. 父进程使用 fork() 创建 x 个子进程
2. 父级创建 x 个管道，每个子级一个
3. 每个子进程写入各自的管道，然后关闭它。
4. 父级从管道读取然后关闭它们。

我目前的草稿尝试如下，但我对其功能有疑问：

// Create x pipes
int fd[2*x];
pipe(fd);

// In child i do (i from 0 to x-1 inclusive):
close(fd[2*i]);
write(fd[2*i +1], .....);
close(fd[2*i +1]);

// In parent do:
wait() // wait for children to finish
// while any pipe has content do:
// For each i from 0 to x-1 inclusive:
close(fd[2*i +1]);
read(fd[2*i], .....);
close(fd[2*i]);

如果有人能向我展示这个概念在工作中的简单示例，我将不胜感激。这里的最终目标是子级与父级之间的一种对话方式，其中父级将存储到单个数组中的多个值。

Answer 1

此代码或多或少可以满足您的需求。它使用我在 GitHub 上的 SOQ（堆栈溢出问题）存储库中提供的错误报告代码，作为 src/libsoq 子目录中的文件 stderr.c 和 stderr.h。

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

enum { NUM_CHILDREN = 5 };
enum { MSG_BUFFSIZE = 64 };
static void be_childish(int kid, int *fd);

int main(int argc, char **argv)
{
    int fd[2 * NUM_CHILDREN];
    int pid[NUM_CHILDREN];

    err_setarg0(argv[0]);
    err_setlogopts(ERR_PID);

    if (argc != 1)
        err_usage("");

    for (int i = 0; i < NUM_CHILDREN; i++)
    {
         if (pipe(&fd[2 * i]) != 0)
             err_syserr("failed to create pipe for child %d: ", i);
    }

    for (int i = 0; i < NUM_CHILDREN; i++)
    {
        if ((pid[i] = fork()) < 0)
            err_syserr("failed to fork child %d: ", i);
        else if (pid[i] == 0)
            be_childish(i, fd);
        else
        {
            printf("Child %d has PID %d\n", i, pid[i]);
            fflush(stdout);
        }
    }

    char buffer[MSG_BUFFSIZE];
    for (int i = 0; i < NUM_CHILDREN; i++)
    {
        close(fd[2 * i + 1]);
        int pipe_in = fd[2 * i + 0];
        int nbytes = read(pipe_in, buffer, sizeof(buffer));
        if (nbytes < 0)
            err_syserr("failed to read from FD %2d: ", pipe_in);
        printf("Got %2d bytes [%.*s] from FD %2d, PID %d\n",
               nbytes, nbytes, buffer, pipe_in, pid[i]);
        close(pipe_in);
    }

    for (int i = 0; i < NUM_CHILDREN; i++)
    {
        int status;
        int corpse = wait(&status);
        if (corpse > 0)
            printf("Child with PID %d exited with status 0x%.4X\n", corpse, status);
        else
            err_syserr("Failed to wait for dead children: ");
    }

    return 0;
}

static void be_childish(int kid, int *fd)
{
    for (int i = 0; i < kid; i++)
    {
        close(fd[2 * i + 0]);
        close(fd[2 * i + 1]);
    }

    close(fd[2 * kid + 0]);
    int estat = kid + 32;

    char buffer[MSG_BUFFSIZE];
    int nbytes = snprintf(buffer, sizeof(buffer),
                          "Child %d (PID %d) exiting with status %d",
                          kid, (int)getpid(), estat);
    int pipe_out = fd[2 * kid + 1];
    if (write(pipe_out, buffer, nbytes) != nbytes)
        err_syserr("failed to write to parent: ");
    close(pipe_out);
    exit(estat);
}

示例运行：

Child 0 has PID 36957
Child 1 has PID 36958
Child 2 has PID 36959
Child 3 has PID 36960
Child 4 has PID 36961
Got 42 bytes [Child 0 (PID 36957) exiting with status 32] from FD  3, PID 36957
Got 42 bytes [Child 1 (PID 36958) exiting with status 33] from FD  5, PID 36958
Got 42 bytes [Child 2 (PID 36959) exiting with status 34] from FD  7, PID 36959
Got 42 bytes [Child 3 (PID 36960) exiting with status 35] from FD  9, PID 36960
Got 42 bytes [Child 4 (PID 36961) exiting with status 36] from FD 11, PID 36961
Child with PID 36960 exited with status 0x2300
Child with PID 36959 exited with status 0x2200
Child with PID 36958 exited with status 0x2100
Child with PID 36957 exited with status 0x2000
Child with PID 36961 exited with status 0x2400