一行中的多个命令的简单shell

Question

我的程序应该打印出一个提示：myshell＆gt;当它准备好接受输入时。它必须读取一行输入，接受几个可能的命令。这是一个非常简单的shell，因此它只接受两个命令：run和exit。

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


 int h=0;
  void keep(int sig){
  h=1;
 }
 int main()
 {
 signal(SIGINT,keep);//gets into the interrupt
    while(1)
 {
 printf("myshell>");
 char line[255];
 fgets(line, 255, stdin);
 if(!strcmp(line,"\n")){//checks if the given line is not empty
    continue;
  }
    if(h)
    {
    h=0;
    continue;
    }
  int g = 0;
  char *words[5];
  words[g] = strtok(line," \n");//Tokenize the given line
   while (words[g]!=NULL)//splitting the tokens
  {
   g++;
   words[g] = strtok(NULL," \n");
   }
  if(strcmp(words[0],"run")==0)
  {
   int status;
   printf("myshell:Started child pid %d \n",getpid());
   pid_t result = fork();
   printf("myshell:Started child pid %d \n",getpid());
   if(result>0)
   {
   wait(&status);//gets into the wait state
   }
   else if  (result == 0)//new child process created
   {
   char *args[3];
   int k=0;
   while(k<4)//take the whole string into array and execute
   {
     args[k] = words[k+1];
     k++;
   }
   int output;
   output = execvp(*args,args);
   if(output==-1)
   {
     printf("myshell: could not find the program %s\n",words[1]);
   }
   }
   }
  else if(strcmp(words[0],"exit")==0)//comparing with the exit
  {
   exit(1);
   }
   else
   {
   printf("myshell: %s is not a valid command\n",words[0]);
   }
   }
   return 0;
   }

它适用于

./myshell
 myshell> run ls
 myshell: started child pid 7973
 myshell myshell.c

我的shell应该支持使用;使用单个命令执行多个程序。为简单起见，我们假设在;之前和之后有一个空格。例如：

 ./myshell
 myshell> run ls -l ; date ; who

但这不适用于此代码。