CODEWITHSUNDEEP
c
user981536
user981536

Reputation: 1

How to use execve for exporting environment variables

I am trying to figure out how to export this environmental variables from my minishell to bash. Nothing happens when I use this code to export. Am I doing something wrong

      if(strcmp(istring, "myexport") ==0)     //This command shows parent enviornment
  {
      char * const *junk;
      execve("/bin/bash" , junk , myexp);
      return(1);
  }

Upvotes: 0

Views: 5137

Answers (3)

Jonathan Leffler
Jonathan Leffler

Reputation: 753695

If the intention is to run bash in order to set the environment in your shell, you are misunderstanding how the environment is set.

There are the coding problems identified in the other answers:

  • You do not initialize junk.
  • You do not show how you initialize myexp.

Assuming you get past those issues, though, the deeper one is that the child process cannot affect the environment of the parent process. When a shell exports an environment variable, it does so by adjusting a list that is supplied to each command it executes, but it is handled inside the shell, not by executing any external command.

Also, if the fragment shown is not run in a child process (after a fork()), then you have another problem; execve() does not return when the command executes successfully.

So, to export an environment variable, the change is made in the main shell, not in a sub-process.

Upvotes: 1

Ed Heal
Ed Heal

Reputation: 59997

Try

char *const argv = { "-c", "env", 0 };
char *const env = { "PATH=/bin", "USER=wibble", 0 };

execve("/bin/bash", argv, env);

And of course you can fetch things our of the existing environment by using extern char **environ; or getenv to construct the environment for the new binary/script.

Upvotes: 2

Adam Rosenfield
Adam Rosenfield

Reputation: 400254

You're passing an uninitialized pointer as the second argument to execve—most likely, that will fail with EFAULT, but if it does somehow magically succeed, bash is going to see garbage in its argv array of arguments and do something strange (most likely complain that some junk file doesn't exist).

If you don't want to pass any arguments, pass in a pointer to NULL, but it's good practice to at least pass in the name of the program being executed as argv[0]:

char *const argv[] = {"bash", NULL};
execve("/bin/bash", argv, myexp);

Also, what is myexp? Is it a pointer to a NULL-terminated array of pointers to strings of the form FOO=BAR?

Upvotes: 0

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