Calling 'ls' with execv

Question

I am new to system calls and C programming and am working on my university assignment.

I want to call the 'ls' command and have it print the directory.

What I have: (I have added comments in so you can see what I see coming through each variable.

int execute( command* cmd ){

  char full_path[50];
  find_fullP(full_path, p_cmd); 
  //find_fullP successfully updates full_path to /bin/ls
  char* args[p_cmd->argc];
  args[0] = p_cmd->name;
  int i;
  for(i = 1; i < p_cmd->argc; i++){
      args[i] = p_cmd->argv[i];
  }

/*
 * this piece of code updates an args variable which holds arguments 
 * (stored in the struct) in case the command is something else that takes 
 * arguments. In this case, it will hold nothing since the command 
 * will be just 'ls'.
 */

  int child_process_status;
  pid_t child_pid;
  pid_t pid;

  child_pid = fork();

  if ( child_pid == 0 ) {
      execv( full_path, args );
      perror("fork child process error condition!" );
  }

  pid = wait( &child_process_status );
  return 0;
}

I am not seeing anything happening and am confused, any idea?

Use `opendir`, `readdir`, and `closedir` instead of using an external program — Ed Heal, Feb 13 '15 at 20:13
Add logging (just printing stuff is fine) to find out how far you get and whether variables hold the values you think they hold. Or use a debugger. — David Schwartz, Feb 13 '15 at 20:16
@EdHeal Do you think that will get full marks when submitted as the university assignment? — Kaz, Feb 13 '15 at 20:17
Your `args` array should terminate with `NULL`. See `man execv` — Eugene Sh., Feb 13 '15 at 20:17
If the assignment is to print a directory, I would expect it to get higher marks. — David Schwartz, Feb 13 '15 at 20:17
Why don't you change the representation of `p_cmd->argv[]` so that `p->cmd_argv[0]` is the program name? Then you don't have to make a copy. You can just use that `argv` object. Don't forget that `argv[argc]` must be a null pointer. The argv for a program with N arguments requires N+2 elements. One for the program name, N for the arguments, and one for the null pointer which terminates the array. If you don't terminate the array, then `/bin/ls` may find a complete garbage pointer in `argv[1]` (so `ls` crashes), or a pointer to junk data (which looks like a file which it cannot find). — Kaz, Feb 13 '15 at 20:18
I added NULL to the end of args and it worked, but now I get an "Abort Trap: 6" error. — user3037172, Feb 13 '15 at 20:21
Using an external program uses fewer system calls. Therefore the use of `opendir` etc would get better marks. (also more efficient) — Ed Heal, Feb 13 '15 at 20:22
Make sure that you have allocated the space for the NULL element. — Eugene Sh., Feb 13 '15 at 20:22
@EdHeal For just printing the directory contents, it's much simpler to use `ls` than messing with own lower level implementation, really. — Eugene Sh., Feb 13 '15 at 20:26
No is not. about 5 lines of code and a format of your choosing. Without the assumption `/bin/ls` or it printing in the desired format — Ed Heal, Feb 13 '15 at 20:29
Perhaps http://stackoverflow.com/questions/3554120/open-directory-using-c is worth a read — Ed Heal, Feb 13 '15 at 20:47

score 14 · Accepted Answer · edited Sep 10 '16 at 04:30

Here's the minimal program that invokes ls using execv. Things to note

the list of args should include the executable as the first arg
the list of args must be NULL terminated
if the args are set up correctly, then args[0] can be passed as the first parameter to execv

#include <sys/types.h>
#include <sys/wait.h>
#include <unistd.h>

int main( void )
{
    int status;
    char *args[2];

    args[0] = "/bin/ls";        // first arg is the full path to the executable
    args[1] = NULL;             // list of args must be NULL terminated

    if ( fork() == 0 )
        execv( args[0], args ); // child: call execv with the path and the args
    else
        wait( &status );        // parent: wait for the child (not really necessary)

    return 0;
}

Calling 'ls' with execv

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