尝试使用循环创建可伸缩的管道和execvp程序

Question

该程序尝试使用多于一个的任意数量的命令，并使用管道、execvp和fork将它们链接在一起，就像shell一样。在这段代码中，我有一个硬编码的"ls“、" wc”和"less“，结果应该像在shell上运行"ls |wc| less”一样。由于某些原因，管道不能按预期工作。我在第99行有一大堆评论来解释我认为的问题所在(以"The read end of the...“开头)。我知道没有错误检查，任何帮助都是非常感谢的。

#include <iostream>
#include <string.h>
#include <string>
#include <vector>
#include <sys/types.h>
#include <unistd.h>
using namespace std;

#define READ 0
#define WRITE 1

//This program will do three different commands ls, wc, then less.
int main(){
  pid_t pid;
  int cmd=3;
  //One less pipe than command is required.
  int fd[cmd-1][2];
  //The pipes are created in a for loop.
  for(int i=0; i<(cmd-1); i++){
    if(pipe(fd[i])==-1){
      cout<<"Help"<<endl;
    }
  }
  //The commands are put in c.
  char* c[3];
  c[0]="ls";
  c[1]="wc";
  c[2]="less";
  //First fork
  pid=fork();
  if(pid==0){

    //Pipe 0 is linked up.
    close(fd[0][READ]);
    dup2(fd[0][WRITE], 1);
    close(fd[0][WRITE]);

    //Remaining pipes are closed.
    for(int i=1; i<(cmd-1); i++){
      close(fd[i][READ]);
      close(fd[i][WRITE]);
    }

    //The command is prepared and then execvp is executed.
    char* temp[2];
    temp[0]=c[0];
    temp[1]=NULL;
    char* x=temp[0];
    execvp(x, temp);
  }

  //This for loop executes two times less than the number of commands.
  for(int i=0; i<(cmd-2); i++){
    pid=fork();
    if(pid==0){

      //I link up the read connection with pipe 0, I am fairly certain that
      //this part is working. You can put a cout after this pipe and it will
      //print that of command 1.
      close(fd[i][WRITE]);
      dup2(fd[i][READ], 0);
      close(fd[i][READ]);

      //This is the linking of pipe 1.
      close(fd[i+1][READ]);
      dup2(fd[i+1][WRITE], 1);
      close(fd[i+1][WRITE]);

      //This closes the remaining pipes, in this case there are none.
      for(int j=0; j<(cmd-1); j++){
    if(j==i || j==(i+1)){
      continue;
    }
    close(fd[j][READ]);
    close(fd[j][WRITE]);
      }

      //The command is prepared and executed
      char* temp[2];
      temp[0]=c[i+1];
      temp[1]=NULL;
      char* x=temp[0];
      execvp(x, temp);
    }
  }
  pid=fork();
  if(pid==0){

    //The read end of the final pipe is linked here.
    //THIS IS WERE THE PROBLEM IS! For some reason after dup2, I can no longer
    //use cin. Inbetween the linking of pipe 0 and pipe 1 (line 66), I can
    //use cin to make sure that the first execvp works and put its output in the
    //pipe. I also know that the second execvp works as intended. I just need to
    //know why dup2 messes up my program here.
    close(fd[cmd-2][WRITE]);
    dup2(fd[cmd-2][READ], 0);
    close(fd[cmd-2][READ]);

    //closes the remaining pipes.
    for(int i=0; i<(cmd-2); i++){
      close(fd[i][READ]);
      close(fd[i][WRITE]);
    }

    //Preps next command.
    char* temp[2];
    temp[0]=c[cmd];
    temp[1]=NULL;
    char* x=temp[0];
    execvp(x, temp);
    //}

  //closes all pipes.
  for(int i=0; i<(cmd-1); i++){
    close(fd[i][READ]);
    close(fd[i][WRITE]);
  }
  return 0;
}

Answer 1

您的代码有多个问题，例如您没有为命令分配内存，您的代码似乎没有正确地包含在括号中

我修改了你的代码如下:

#include <iostream>
#include <string.h>
#include <string>
#include <vector>
#include <sys/types.h>
#include <unistd.h>
using namespace std;


//This program will do three different commands ls, wc, then less.
int main(){
  pid_t pid = 0;
  int cmd=3, i;
  //One less pipe than command is required.
  int fd[cmd-1][2];
  //The pipes are created in a for loop.
  for(int i=0; i<(cmd-1); i++){
    if(pipe(fd[i])==-1){
      cout<<"Help"<<endl;
    }
  }
  //The commands are put in c.
  char c[3][8] = {{'l', 's', '\0'}, {'w', 'c', '\0'}, {'l','e','s','s', '\0'}}, *temp[2];

  for(i = 0; i < cmd-1; i ++){
    pid = fork();
    if(pid == 0){
      if(i != 0){
        // read from previous fd
        close(fd[i-1][1]);
        dup2(fd[i-1][0], STDIN_FILENO);
        close(fd[i-1][0]);
      }
      // write to current fd
      close(fd[i][0]);
      dup2(fd[i][1], STDOUT_FILENO);
      close(fd[i][1]);
      temp[0] = c[i];
      temp[1] = NULL;
      execvp(c[i], temp);
      exit(0);
    }
    else{
      if(i != 0){
        // close unnecessary fds in parent
        close(fd[i-1][0]);
        close(fd[i-1][1]);
      }
    }
  }
  // the last command i.e. less here
  if(i > 0){
      close(fd[i-1][1]);
      dup2(fd[i-1][0], STDIN_FILENO);
      close(fd[i-1][0]);
  }
  temp[0] = c[i];
  temp[1] = NULL;
  execvp(c[i], temp);
  return 0;
}

如果对你有效，请让我知道！