strtok and execlp in a mini-shell

Question

Hi fellows!

I'm writing a mini-shell to get more familiar with Unix process management in C. It's reading stuff from commandline and passes these arguments via execlp to the system.

# include <stdio.h>
# include <stdlib.h>
# include <unistd.h>

#define MAXSIZE 100

char prompt[MAXSIZE];

int 
main(void){


   pid_t pid;

   printf("> ");

   // read stuff 
   if (fgets(prompt, MAXSIZE, stdin) == NULL){
      printf("Input validation error!");
      abort();
   }
   // printf("DEBUG: %s" , prompt);

   if (strcmp(prompt, "exit")==0) abort();


   if ((pid=fork())<0){       // copy process

      printf("Process error!");
      abort();
   }

   if (pid==0){                // exec in son-prcess

      char *command=(char*)strtok(prompt, " ");
      execlp(command, command, 0);  // overwrite memory

      printf("Error, command not found!");
      abort();
   } else {

      waitpid(pid, 0, 0);
    }
}

In fact this would be it... but I don't get any output from execlp. Does anybody know why that is?

Thanks ;), Marius

Answer 1

I tried running your program and it failed because command contained a \n (newline). I altered it by putting \n instead of " " in the strtok and it then ran successfully.

In detail:

  if (pid==0){                // exec in son-prcess
      char *command=(char*)strtok(prompt, "\n");
      printf ("'%s'\n", command);
      execlp (command, command, 0);  // overwrite memory
      printf("Error %d (%s)\n", errno, strerror (errno));
      abort();
   } else {

Test run:

$ ./a.out 
> ls
'ls'
(usual ls behaviour)

Answer 2

Kinopiko already found why it doesn't work, but the reason you didn't see any error message is that your shell prompt is overwriting it. Try putting a newline at the end:

printf("Error, command not found!\n");