I am using while loop to enter characters one by one(total number of characters to be entered is unknown). How can I store my characters in an array:

Question

User has to input a string of unknown length(<1000). So here I am using while loop, #define, getchar.
What should I do to store the characters simultaneously?

#include <stdio.h>
#define EOL '\n'

int main()
{
    int count=0;
    char c;

    while((c=getchar())!= EOL)// These c's have to be stored.
    {
        count++;
    }

    return 0;
}

EDIT: Sorry I didn't tell this thing earlier. I don't have to waste 1000 bytes if count!=1000. That's why I am not declaring any array of 1000 elements initially.

Answer 1

Implement a dynamically growing array

1. use calloc() or malloc() to allocate a small memory block - let's say for 10 chars

2. use realloc() to increase the size of the memory block if needed

Example:

#include <stdio.h>
#include <stdlib.h>

#define EOL '\n'

int main()
{
    int count = 0;
    int size = 10;
    char *chars = NULL;
    char c;
    
    /* allocate memory for 10 chars */
    chars = calloc(size, sizeof(c));
    if(chars == NULL) {
        printf("allocating memory failed!\n");
        return EXIT_FAILURE;
    }

    /* read chars ... */
    while((c = getchar()) != EOL) {
        /* re-allocate memory for another 10 chars if needed */
        if (count >= size) {
            size += size;
            chars = realloc(chars, size * sizeof(c));
            if(chars == NULL) {
                printf("re-allocating memory failed!\n");
                return EXIT_FAILURE;
            }
        }
        chars[count++] = c;
    }

    printf("chars: %s\n", chars);

    return EXIT_SUCCESS;
}