6

这个问题源于我尝试执行以下指令:

Linux 管道作为输入和输出

如何使用管道在两个程序之间发送一个简单的字符串?

http://tldp.org/LDP/lpg/node11.html

我的问题类似于:Linux Pipes as Input and Output中的问题,但更具体。

本质上,我正在尝试替换:

/directory/program < input.txt > output.txt

在 C++ 中使用管道以避免使用硬盘驱动器。这是我的代码:

//LET THE PLUMBING BEGIN 
int fd_p2c[2], fd_pFc[2], bytes_read;
    // "p2c" = pipe_to_child, "pFc" = pipe_from_child (see above link)
pid_t childpid;
char readbuffer[80];
string program_name;// <---- includes program name + full path
string gulp_command;// <---- includes my line-by-line stdin for program execution
string receive_output = "";

pipe(fd_p2c);//create pipe-to-child
pipe(fd_pFc);//create pipe-from-child
childpid = fork();//create fork

if (childpid < 0)
{
    cout << "Fork failed" << endl;
    exit(-1);
}
else if (childpid == 0)
{
    dup2(0,fd_p2c[0]);//close stdout & make read end of p2c into stdout
    close(fd_p2c[0]);//close read end of p2c
    close(fd_p2c[1]);//close write end of p2c
    dup2(1,fd_pFc[1]);//close stdin & make read end of pFc into stdin
    close(fd_pFc[1]);//close write end of pFc
    close(fd_pFc[0]);//close read end of pFc

    //Execute the required program
    execl(program_name.c_str(),program_name.c_str(),(char *) 0);
    exit(0);
}
else
{
    close(fd_p2c[0]);//close read end of p2c
    close(fd_pFc[1]);//close write end of pFc

    //"Loop" - send all data to child on write end of p2c
    write(fd_p2c[1], gulp_command.c_str(), (strlen(gulp_command.c_str())));
    close(fd_p2c[1]);//close write end of p2c

    //Loop - receive all data to child on read end of pFc
    while (1)
    {        
        bytes_read = read(fd_pFc[0], readbuffer, sizeof(readbuffer));

        if (bytes_read <= 0)//if nothing read from buffer...
            break;//...break loop

        receive_output += readbuffer;//append data to string
    }
    close(fd_pFc[0]);//close read end of pFc
}

我绝对确定上述字符串已正确初始化。但是,发生了两件事对我来说没有意义:

(1) 我正在执行的程序报告“输入文件为空”。由于我没有使用“<”调用程序,因此不应期待输入文件。相反,它应该期待键盘输入。此外,它应该阅读“gulp_command”中包含的文本。

(2) 程序的报告(通过标准输出提供)出现在终端中。这很奇怪,因为此管道的目的是将标准输出传输到我的字符串“receive_output”。但由于它出现在屏幕上,这向我表明信息没有通过管道正确传递到变量。如果我在 if 语句的末尾执行以下操作,

cout << receive_output << endl;

我什么也没得到,好像字符串是空的。我很感激你能给我的任何帮助!

编辑:澄清

我的程序当前使用文本文件与另一个程序通信。我的程序编写了一个文本文件(例如 input.txt),由外部程序读取。然后该程序生成 output.txt,由我的程序读取。所以它是这样的:

my code -> input.txt -> program -> output.txt -> my code

因此,我的代码目前使用,

system("program < input.txt > output.txt");

我想用管道替换这个过程。我想将我的输入作为标准输入传递给程序,并让我的代码将该程序的标准输出读入一个字符串。

4

3 回答 3

7

您的主要问题是您的论点要dup2()颠倒。你需要使用:

dup2(fd_p2c[0], 0);   // Duplicate read end of pipe to standard input
dup2(fd_pFc[1], 1);   // Duplicate write end of pipe to standard output

在我对设置代码进行错误检查并从调用中得到意外值之前,我被误读为 OK dup2(),这告诉我问题出在哪里。当出现问题时,插入您之前跳过的错误检查。

您也没有确保从孩子读取的数据为空终止;这段代码可以。

工作代码(带诊断),使用cat最简单的“其他命令”:

#include <unistd.h>
#include <string>
#include <iostream>
using namespace std;

int main()
{
    int fd_p2c[2], fd_c2p[2], bytes_read;
    pid_t childpid;
    char readbuffer[80];
    string program_name = "/bin/cat";
    string gulp_command = "this is the command data sent to the child cat (kitten?)";
    string receive_output = "";

    if (pipe(fd_p2c) != 0 || pipe(fd_c2p) != 0)
    {
        cerr << "Failed to pipe\n";
        exit(1);
    }
    childpid = fork();

    if (childpid < 0)
    {
        cout << "Fork failed" << endl;
        exit(-1);
    }
    else if (childpid == 0)
    {
        if (dup2(fd_p2c[0], 0) != 0 ||
            close(fd_p2c[0]) != 0 ||
            close(fd_p2c[1]) != 0)
        {
            cerr << "Child: failed to set up standard input\n";
            exit(1);
        }
        if (dup2(fd_c2p[1], 1) != 1 ||
            close(fd_c2p[1]) != 0 ||
            close(fd_c2p[0]) != 0)
        {
            cerr << "Child: failed to set up standard output\n";
            exit(1);
        }

        execl(program_name.c_str(), program_name.c_str(), (char *) 0);
        cerr << "Failed to execute " << program_name << endl;
        exit(1);
    }
    else
    {
        close(fd_p2c[0]);
        close(fd_c2p[1]);

        cout << "Writing to child: <<" << gulp_command << ">>" << endl;
        int nbytes = gulp_command.length();
        if (write(fd_p2c[1], gulp_command.c_str(), nbytes) != nbytes)
        {
            cerr << "Parent: short write to child\n";
            exit(1);
        }
        close(fd_p2c[1]);

        while (1)
        {
            bytes_read = read(fd_c2p[0], readbuffer, sizeof(readbuffer)-1);

            if (bytes_read <= 0)
                break;

            readbuffer[bytes_read] = '\0';
            receive_output += readbuffer;
        }
        close(fd_c2p[0]);
        cout << "From child: <<" << receive_output << ">>" << endl;
    }
    return 0;
}

样本输出:

Writing to child: <<this is the command data sent to the child cat (kitten?)>>
From child: <<this is the command data sent to the child cat (kitten?)>>

请注意,您需要小心确保您的代码不会陷入僵局。如果你有一个严格同步的协议(所以父母写一条消息并在锁步中读取一个响应),你应该没问题,但是如果父母试图写一条太大而无法放入管道的消息给孩子当孩子试图将一条太大而无法放入管道中的消息返回给父母时,每个人都会在等待另一个人阅读时被阻止写入。

于 2013-07-07T15:24:20.237 回答
1

听起来您正在寻找coprocesses。您可以在 C/C++ 中对它们进行编程,但由于它们已经在 (bash) shell 中可用,因此更易于使用 shell,对吧?

首先使用coproc内置启动外部程序:

coproc external_program

coproc后台启动程序并将文件描述符存储在数组 shell 变量中以与之通信。现在您只需要启动程序将其连接到这些文件描述符:

your_program <&${COPROC[0]} >&${COPROC[1]}
于 2013-07-07T09:17:41.050 回答
1
#include <stdio.h>
#include <unistd.h>
#include <sys/stat.h>
#include <sys/wait.h>
#include <fcntl.h>
#include <string.h>
#include <iostream>
using namespace std;
int main() {
    int i, status, len;
    char str[10];
    mknod("pipe", S_IFIFO | S_IRUSR | S_IWUSR, 0); //create named pipe
    pid_t pid = fork(); // create new process
    /* Process A */
    if (pid == 0) {
        int myPipe = open("pipe", O_WRONLY); // returns a file descriptor for the pipe
        cout << "\nThis is process A having PID= " << getpid(); //Get pid of process A
        cout << "\nEnter the string: ";
        cin >> str;
        len = strlen(str);
        write(myPipe, str, len); //Process A write to the named pipe
        cout << "Process A sent " << str;
        close(myPipe); //closes the file descriptor fields.
        }
    /* Process B */
        else {
        int myPipe = open("pipe", O_RDONLY); //Open the pipe and returns file descriptor
        char buffer[21];
        int pid_child;
        pid_child = wait(&status); //wait until any one child process terminates
        int length = read(myPipe, buffer, 20); //reads up to size bytes from pipe with descriptor fields, store results
    //  in buffer;
        cout<< "\n\nThis is process B having PID= " << getpid();//Get pid of process B
        buffer[length] = '\0';
        cout << "\nProcess B received " << buffer;
        i = 0;
        //Reverse the string
        for (length = length - 1; length >= 0; length--)
        str[i++] = buffer[length];
        str[i] = '\0';
        cout << "\nRevers of string is " << str;
        close(myPipe);
        }
    unlink("pipe");
return 0;
}
于 2014-10-15T10:08:46.633 回答