我不会撒谎.这是一个家庭作业问题.但是,据我所知,这些点已经消失了.现在,我只是在寻找答案,因为我想 – 我可能会疯了.

该程序的目标是执行命令ps -A | grep(inputstring)| wc -l的方式类似于shell的方式.所以,我产生了进程,让它们互相等待.最新的进程,曾孙,execlp(“ps”,“ps”,“ – A”,NULL),用ps -A进程取而代之.在execlp之前,我确保它的标准输出转到管道输出.行中的下一个进程是wait()ing,并且已经自己设置,以便输入管道进入标准输入,标准输出进入输出管道,它将执行grep,依此类推.

我几乎是肯定的,我已经正确设置了它.然而……程序确实如此.不.工作.

#include <stdlib.h>
#include <iostream>
#include <string>

#define MAXLINE 1500
#define READ 0
#define WRITE 1

using namespace std;

int main( int argc, char** argv ) {
//* start of input block
if ( argc != 2 ) {
    cout << "Usage: ./a.out arg1" << endl;
    return 0;
}
string in = argv[1];
// end of input block */
int pipeA[2], pipeB[2], pid, stat;

// get our first set of pipes
if ( pipe(pipeA) < 0 ) {
    cerr << "Pipe error.\n";
    exit(-1);
}
if ( pipe(pipeB) < 0 ) {
    cerr << "Pipe error.\n";
    exit(-1);
}

// make the first fork
if ( (pid = fork() ) < 0 ) { cerr << "Fork error.\n"; exit(-1); }

if ( pid > 0 ) {    // parent case
    wait(&stat);
} else {            // child case
    if ( (pid = fork()) < 0 ) { cerr << "Fork Error\n"; exit(-1); }
    if ( pid > 0 ) {    // child
        wait(&stat);
        dup2(pipeA[READ],READ);
        execlp("wc","wc","-l",NULL);
    } else {    // grand-child
        if ( (pid = fork()) < 0 ) { cerr << "Fork Error\n"; exit(-1); }
        if ( pid > 0 ) {    // still grand-child
            wait(&stat);
            dup2(pipeB[READ],READ);  
            dup2(pipeA[WRITE],WRITE); 
            close(pipeB[READ]);
            execlp("grep","grep",in.c_str(),NULL);
        } else {    // great grand-child
            dup2(pipeB[WRITE],WRITE); // t now goes to pipeB[1]
            close(READ);
            close(pipeB[READ]);
            execlp("ps", "ps", "-A", NULL);
        }
    }
}
return 0;
}

编辑：更改为我的代码的双管变体.

解决方法:

我几乎可以肯定这就是你要做的事情.为马虎编码提前道歉.它在这里有点晚了,我现在应该睡觉了：

#include <iostream>
#include <string>
#include <cstdlib>
#include <cstdio>
#include <unistd.h>

#define READ 0
#define WRITE 1

// ps -A | grep argv[1] | wc -l

int main( int argc, char** argv )
{
    //* start of input block
    if ( argc != 2 )
    {
        std::cout << "Usage: ./a.out arg1" << std::endl;
        return 0;
    }

    // make local copy of argument
    std::string in = argv[1];
    int fd1[2], fd2[2], pid;

    // allocate two pipe sets
    if (pipe(fd1) < 0 || pipe(fd2) < 0)
    {
        perror("Failed to create pipe.");
        return EXIT_FAILURE;
    }

    // launch first child process.
    if ((pid = fork()) < 0)
    {
        perror("Failed to fork child(1)");
        return EXIT_FAILURE;
    }

    if (pid == 0)
    {
        // wc -l process. 
        //  stdin  = fd2(read)
        close(fd1[READ]);
        close(fd1[WRITE]);
        close(fd2[WRITE]);
        dup2(fd2[READ],STDIN_FILENO);
        execlp("wc","wc","-l",NULL);
    }

    // fork again. this time for grep
    if ((pid = fork()) < 0)
    {
        perror("Failed to fork child(2)");
        return EXIT_FAILURE;
    }

    if (pid == 0)
    {
        // grep argv[1] process.
        //  stdin  = fd1(read)
        //  stdout = fd2(write)            
        close(fd1[WRITE]);
        close(fd2[READ]);
        dup2(fd2[WRITE], STDOUT_FILENO);
        dup2(fd1[READ], STDIN_FILENO);
        execlp("grep", "grep", in.c_str(), NULL);
    }

    //  fork once more. this time for ps -A
    if ((pid = fork()) < 0)
    {
        perror("Failed to fork child(3)");
        return EXIT_FAILURE;
    }

    if (pid == 0)
    {
        // ps -A process.
        //  stdout = fd1(write)
        close(fd2[WRITE]);
        close(fd2[READ]);
        close(fd1[READ]);
        dup2(fd1[WRITE], STDOUT_FILENO);
        execlp("ps", "ps", "-A", NULL);
    }

    int stat=0;
    wait(&stat);

    return EXIT_SUCCESS;
}

在我的系统上,ps -A报告了141行,其中41行中有单词System some within,只需运行ps -A | grep系统| wc -l.上面的代码生成完全相同的输出.

希望它可以帮助你.