C - How to copy part of an array of string into another?

Question

Is there a simple and elegant way to split an array from an index?

In my program I am getting a array of strings (argv), and I want to ignore the program name and some of its arguments, copying the rest in an array.

For example, the content of argv is {"program_name", "-o", "file1", "file2"}

I want to retrieve "file1" and "file2" in an array just to make an easy iteration over it.

 // PSEUDO
 char *files[argc - 2] = argv.split(2, argc - 2)

Any ideas?

Hmm, argv is not usually a vector. Please make a minimal reproducible example — cigien
– cigien, Commented Oct 11, 2020 at 18:17
why don't you just loop over argv starting from the desired index — Saadi Toumi Fouad
– Saadi Toumi Fouad, Commented Oct 11, 2020 at 18:18
@cigien he probably just meant "array" and used the wrong word. Anyway, there really isn't a better way other than iterating with a loop. You could also just make a loop from i=2 to i<argc, there is no need to copy. — Leonardo Gazdek
– Leonardo Gazdek, Commented Oct 11, 2020 at 18:18
In my code, if no files given, I will have to work with a default. Just to make something generalized. — Raul
– Raul, Commented Oct 11, 2020 at 18:19
C or C++? The answers for each are quite different, please only tag one of the languages — Alan Birtles
– Alan Birtles, Commented Oct 11, 2020 at 18:19

Tiago Loureiro · Accepted Answer · 2020-10-11 18:37:29Z

3

You can use pointer arithmetic to make it in an elegant way:

char** filenames = argv + 2;

Just be sure that you have at least 2 arguments before your filenames as you wish.

#include <stdio.h>

int main(int argc, char** argv) {
  if(argc > 2){
      //char** that points to the first filename
      char** filenames = argv + 2;
      //number of filenames available to iterate
      int num_of_filenames = argc - 2;

      //Printing each name
      int i = 0;
      for(i = 0; i < num_of_filenames; i++){
        printf("%s\n",filenames[i]);
      }
    }
  return 0;
}

edited Oct 11, 2020 at 18:37

answered Oct 11, 2020 at 18:26

Tiago Loureiro

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