使用dup2使C程序执行诸如'ls / bin | grep | grep | grep b'

Question

我在使用dup2使ac程序执行ls /bin | grep grep | grep b等命令时遇到问题 ls /bin | grep grep | grep b ls /bin | grep grep | grep b 。 当我注释掉第三个命令和相关管道时，它执行ls /bin | grep grep ls /bin | grep grep很好，但是使用最后一个命令，它会立即返回。 同样，当我输入“ ps”时，进程仍在运行。 我认为这是由于我如何关闭管道。 我的代码如下：

int main()
{
    int pipeA[2];
    int pipeB[2];

    pipe(pipeA);
    pipe(pipeB);

    int pidA,pidB,pidC;

    if(pidA = fork())
    {
            close(pipeA[0]);
            dup2(pipeA[1],1);
            close(pipeA[1]);
            execlp("ls","ls","/bin",NULL);
            printf("error\n");
    }

    if(pidB = fork())
    {
            close(pipeA[1]);
            dup2(pipeA[0],0);
            close(pipeA[0]);

            close(pipeB[0]);
            dup2(pipeB[1],1);
            close(pipeB[1]);
            execlp("grep","grep","grep",NULL);
            printf("error\n");
    }

    if(pidC = fork())
    {
            close(pipeB[1]);
            dup2(pipeB[0],0);
            close(pipeB[0]);
            execlp("grep","grep","b",NULL);
            printf("error");
    }


    while(pidA != wait(0)){}

    return 0;
}

Answer 1

您没有关闭足够的文件描述符。

/* Semi-working code */
int main()
{
    int pipeA[2];
    int pipeB[2];

    pipe(pipeA);
    pipe(pipeB);

    int pidA,pidB,pidC;

    if (pidA = fork())
    {
            close(pipeB[0]);  // "ls" is not going to use the second pipe
            close(pipeB[1]);  // Ditto
            close(pipeA[0]);
            dup2(pipeA[1], 1);
            close(pipeA[1]);
            execlp("ls", "ls", "/bin", (char *)NULL);
            fprintf(stderr, "error executing 'ls'\n");
            exit(1);
    }

    if (pidB = fork())
    {
            close(pipeA[1]);
            dup2(pipeA[0],0);
            close(pipeA[0]);
            close(pipeB[0]);
            dup2(pipeB[1],1);
            close(pipeB[1]);
            execlp("grep", "grep", "grep", (char *)NULL);
            fprintf(stderr, "error execing 'grep grep'\n");
            exit(1);
    }

    if (pidC = fork())
    {
            close(pipeA[0]);  // The second grep is not going to use the first pipe
            close(pipeA[1]);  // Ditto
            close(pipeB[1]);
            dup2(pipeB[0],0);
            close(pipeB[0]);
            execlp("grep", "grep", "b", (char *)NULL);
            fprintf(stderr, "error execing 'grep b'\n");
            exit(1);
    }

    close(pipeA[0]);  // The parent process is not using the pipes at all
    close(pipeA[1]);
    close(pipeB[0]);
    close(pipeB[1]);

    while (pidA != wait(0))
        ;

    return 0; 
}

因为您没有在第二个grep关闭pipeA ，所以最后一个第一个grep等待来自第二个grep仍然打开的管道的输入，即使该进程不会对其进行写入。 因此，第一个grep不会完成，因此第二个grep也不会完成-即使ls已经完成。 即使父进程关闭了其管道副本，这些注释也将适用-就像更正后的代码一样。

请注意，如何最终关闭四个进程中的每个进程（三个子进程和父进程pipe()的两次调用pipe()返回的所有4个描述符。

这留下了一个残留的问题-由于您常规使用if (pidA = fork())因此流程层次结构颠倒了。 您有一个子进程正在等待其父进程。 您需要使用：

if ((pidA = fork()) == 0)
{
    /* Be childish */
}

其他两个过程中的每个过程类似。 为了确保安全，您还应该检查pipe()调用和fork()调用是否失败。

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

static void err_exit(const char *format, ...);

/* Working code */
int main(void)
{
    int pipeA[2];
    int pipeB[2];

    if (pipe(pipeA) != 0 || pipe(pipeB) != 0)
        err_exit("Failed to create a pipe\n");

    int pidA,pidB,pidC;

    if ((pidA = fork()) < 0)
        err_exit("Failed to fork (A)\n");
    else if (pidA == 0)
    {
            close(pipeB[0]);  // "ls" is not going to use the second pipe
            close(pipeB[1]);  // Ditto
            close(pipeA[0]);
            dup2(pipeA[1], 1);
            close(pipeA[1]);
            execlp("ls", "ls", "/bin", (char *)NULL);
            err_exit("error executing 'ls'\n");
    }

    if ((pidB = fork()) < 0)
        err_exit("failed to fork (B)\n");
    else if (pidB == 0)
    {
            close(pipeA[1]);
            dup2(pipeA[0],0);
            close(pipeA[0]);
            close(pipeB[0]);
            dup2(pipeB[1],1);
            close(pipeB[1]);
            execlp("grep", "grep", "grep", (char *)NULL);
            err_exit("error execing 'grep grep'\n");
    }

    if ((pidC = fork()) < 0)
        err_exit("failed to fork (C)\n");
    else if (pidC == 0)
    {
            close(pipeA[0]);  // The second grep is not going to use the first pipe
            close(pipeA[1]);  // Ditto
            close(pipeB[1]);
            dup2(pipeB[0],0);
            close(pipeB[0]);
            execlp("grep", "grep", "b", (char *)NULL);
            err_exit("error execing 'grep b'\n");
    }

    close(pipeA[0]);  // The parent process is not using the pipes at all
    close(pipeA[1]);
    close(pipeB[0]);
    close(pipeB[1]);

    while (wait(0) != -1)
        ;

    printf("Continuing here...\n");
    sleep(3);
    printf("That's enough of that!\n");

    return 0; 
}

static void err_exit(const char *format, ...)
{
    va_list args;
    va_start(args, format);
    vfprintf(stderr, format, args);
    va_end(args);
    exit(1);
}

当尝试使用/usr/bin代替/bin ，此程序在Mac OS X 10.7.3上可以正常运行。 它列出了三个文件，然后生成有关“在此处继续”的消息：

bzegrep
bzfgrep
bzgrep
Continuing here...
That's enough of that!