前叉：使用一個arg運行shell時，資源暫時不可用

Question

我正在嘗試用C ++寫一個微型外殼，它將帶有1或2個args並在UNIX中運行它們。 我的外殼使用兩個||分隔的參數 很好，但是當我只運行一個時，我得到了一個很大的fork錯誤。 我的外殼會尋找|| 作為管道，而不僅僅是|。 先感謝您！

一些功能命令是：

貓文件名|| 分類

ls -l || 減

碼：

#include <iomanip>
#include <iostream>
#include <string.h>
#include <unistd.h>
#include <stdlib.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <errno.h>
#include <stdio.h>

using namespace std;

void getParms (char[], char* [], char* []); 

int main() 
{
  char command[160];
  pid_t pid1 = 1, pid2 = 1; 

  cout << "myshell> ";
  cin.getline(command, 160);

  while (strcmp(command, "q") != 0 && strcmp(command, "quit") != 0 && pid1 > 0 && pid2 > 0)
   {
    char* arg1[6];    
    char* arg2[6];   
    char path1[21], path2[21];  
    int pipefd[2]; 
    arg1[0]=NULL;
    arg2[0]=NULL;

    getParms(command, arg1, arg2); 

    if (pipe(pipefd) < 0) 
    {
      perror ("Pipe");
      exit (-1);
    } 

//cerr <<"This is arg2"<<arg2[0]<<endl;
    pid1 = fork();
    if (pid1 < 0) 
    {
      perror ("Fork");
      exit (-1);
    }

    if (pid1 == 0) 
    { 
//cout<<"Child 1"<<endl;
//cerr<<arg1[0]<<endl;

      if(arg2[0] != NULL)
      {
      close(pipefd[0]); 
      close(1);      
      dup(pipefd[1]); 
      close(pipefd[1]);
      }

      strcpy(path1, "/bin/"); 
      strcat(path1, arg1[0]);
      if (execvp(path1, arg1) < 0)
      { 
        strcpy(path1, "/usr/bin/"); 
        strncat(path1, arg1[0], strlen(arg1[0]));
        if (execvp(path1, arg1) < 0)
        {
          cout<<"Couldn't execute "<<arg1[0]<<endl; 
          exit (127);
        }
      }

      if(arg2[0]== NULL)
      { // Parent process
        close (pipefd[0]); //read
        close (pipefd[1]); //write

        waitpid(pid1, NULL, 0); // Waits for child2   
        cout << "myshell> ";
        cin.getline(command, 160);
      }
    }

    else if(arg2[0] != NULL)
    { 
//cerr<<"Child 2"<<endl;
      pid2 = fork();
      if (pid2 < 0)
      {
        perror ("Fork");
        exit (-1);
      }

      if (pid2 == 0) 
      { 
        close(pipefd[1]);
        close(0); 

        dup(pipefd[0]);
        close(pipefd[0]);


        strcpy(path2, "/bin/");
        strncat(path2, arg2[0], strlen(arg2[0]));
        if (execvp(path2, arg2) < 0) 
        { 
          strcpy(path2, "/usr/bin/");
          strncat(path2, arg2[0], strlen(arg2[0]));
          if (execvp(path2, arg2) < 0) 
          { 
            cout<<"Couldn't execute "<<arg2[0]<<endl;
            exit (127);
          }
        }
      }

      else 
      { // Parent process
//cerr<<"in last 2 else"<<endl;
        close (pipefd[0]); //read
        close (pipefd[1]); //write

        waitpid(pid2, NULL, 0); // Waits for child2   
        cout << "myshell> ";
        cin.getline(command, 160);
      }
    }
  }
  return 0;
 }


/****************************************************************
   FUNCTION:   void getParms (char [], char* [], char* [])

   ARGUMENTS:  char str[] which holds full command 
               char* args[] args2[] which will hold the individual commands

   RETURNS: N/A

****************************************************************/
void getParms(char str[], char* args[], char* args2[])
 {
  char* index;
  int i= 0; 
  int j= 0; 

  index = strtok(str, " ");
//cerr<<"before first while"<<endl;

  // While the token isn't NULL or pipe
  while (index != NULL && strstr(index,"||") == NULL)
   {
    args[i] = index;
    index = strtok(NULL, " ");
    i++;
   }
   args[i] = (char*) NULL; // makes last element Null 

//cerr<<" getParms before ||"<<endl;
   if(index != NULL && strcmp(index,"||") != 0)
   {
//cerr<<"after checking for ||"<<endl;
      index = strtok(NULL," ");
      while (index != NULL)
      {
        args2[j] = index;
        index = strtok(NULL," ");
        j++;
      }
   }
//cerr<<"After second IF"<<endl;
   args2[j] = (char*) NULL; // makes last element Null
 }

Answer 1

您的問題是，主while循環不會進入任何提示您輸入其他命令的if-else語句-一次又一次地執行同一條語句。 當您使用雙管道時，轉到else if(arg2[0] != NULL) ，並且父進程顯示新的提示。

嘗試從if-else語句的主while循環中刪除命令的兩個提示，然后將提示移動到循環的開頭，如下所示：

//Move these two below into the while loop
//cout << "myshell> ";
//cin.getline(command, 160);

while (strcmp(command, "q") != 0 && strcmp(command, "quit") != 0 && pid1 > 0 && pid2 > 0)
{
    cout << "myshell> ";
    cin.getline(command, 160);
    //...
}

盡量不要對同一事物進行此類冗余調用。 如果您有幾個，並且您需要更改某些內容，則可能會變得凌亂。