How to pass arguments from terminal to a function

Question

Possible Duplicate:
Pass arguments into C program from command line

I am trying to pass three arguments from terminal into a function called replace. I would like to know if it is possible to do the following from terminal

  % ./replace d DDD mytest.tx

I have looked online but can only find information on passing values directly to main() and not the function inside.

Edit: I have edited the main functions as following:

void replace(char* string_a, char* string_b, char* string_f)
 {
  }

int main(int argc, char *argv[])
  { 
       if(argc < 4)
  { 
    printf("Not enough arguments\n");
   return 0;
  }

replace(argv[1],argv[2],argv[3]);
 }

Solution 1

Your main function should be in the format:

int main ( int argc, char *argv[] )

argv is a pointer to your arguments. Note that the first argument is the name of your program.

Here is a lesson on command line arguments:

Solution 2

Course you can, main prototype is int main(int ac, char **av) where ac is the number of arguments passed to the program and char** is an array of arrays containing the arguments passed to the program.

For example for your code :

int main(int ac, char **av)
     {
      void replace (av[1], av[2], av[3])
        { 
         .......
        }

     }

If you launch your exec as : ./replace d DDD mytest.tx, av[0] will be your program name, av[1] will be d, av[2] DDD and av[3] mytest.tx

Good luck !

Solution 3

All stand-alone proper C programs start their execution in main(). That's just how the language works.

So what you need to do is call replace() from main(), after verifying and interpreting the arguments. The command-line arguments will be in the standard int argc, char *argv[] parameters to main().

Solution 4

The thing you want to do is use main's argument list.

You can use the following signature:

int main(int argc, char* argv[]), where argv is a pointer to the argument list, passed from the command line.

Solution 5

Look up argc and argv.

argc is the number of arguments and argv is an array of pointers to your arguments.

so your code should look something like this:

void replace (char string_a[],char string_b[], char string_f[])
{
    //...
}

int main(int argc, char *argv[])
{
    if(argc < 4)
    {
        printf("Not enough arguments\n");
        return 0;
    }

    replace(argv[1], argv[2], argv[3]);
}

Also remember that the first item in argv (argv[0]) is the path of the executing program.

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Comments

[Mike]

Why not pass them to main() then to the function from there? Why are you trying to by-pass sending them to main()?

[Tomislav Dyulgerov]

Actually av[0] will be the name of the program as Lews Therin has correctly pointed out.

[Simon MILHAU]

Yep mistaken this =S Edited !

[user1072706]

Thanks for the explanation. No, I am not trying to bypass main; my goal is simply to call replace from terminal.