Storing characters in array using pointers - c

Question

I am supposed to create an array with the fixed size (let’s say 5x5). I am supposed to fill it with words typed as command line parameters. I can use only pointers. Obviously I have to create two dimensional array[5][5]. I can’t use more than 25 letters ( I will make if statements ). I am going replace white space with some symbol ($ or something). The problem is I don’t know how to connect those command line parameters with array using only pointers.

Answer 1

You need to use the argc and argv arguments from the main function.
Here is an example.
#include <stdio.h>
int main(int argc, char *argv[])
{
for (unsigned i = 1; i < argc; i++)
{
printf("Word %u is: %s\n", i, argv[i]);
}
return 0;
}
You compile it using gcc main.c -o a.exe.
And run it using a.exe hello world.
Edit:
Further to your comment.
You can create an array of pointers like so:
#include <stdio.h>
#include <string.h>
#include <stddef.h>
#include <stdlib.h>
int main(int argc, char *argv[])
{
char *my_array_of_strings[5]; // up to 5 strings
// populate your array
for (unsigned i = 1; i < argc; i++)
{
my_array_of_strings[i - 1] = malloc(strlen(argv[i]) + 1);
if (NULL == my_array_of_strings[i - 1])
{
printf("oh no, memory failure");
break;
}
memcpy(my_array_of_strings[i - 1], argv[i], strlen(argv[i]) + 1);
printf("word is %s\n", my_array_of_strings[i - 1]);
}
return 0;
}