how to make a copy of string into a dynamically allocated character array of the appropriate size?

Question

char input[1000];

I want to copy input into a dynamically allocated character array, how do I approach this problem.

So far I have used strncpy, but get lots of errors.

Answer 1

Are you looking for something like this:

#include <stdlib.h>
#include <stdio.h>
#include <string.h>    
    int main() {
        int i;
        char input[1000] = "Sample string";
        char *in = malloc(1000 * sizeof(char)); // use dynamic number instead of 1000
        strcpy(in, input);
        for (i = 0; i < 5; ++i) { // intentionally printing the first 5 character
            printf("%c", in[i]);
        }
    }

The output is:

Sampl

Edit: In C++ the cast is required for malloc, so I write:

(char *)malloc(1000 * sizeof(char))

But in C, never cast the result of malloc().

Answer 2

What you can do is just use strcpy() offer by C string.h after dynamically allocating memory to your array, as shown:

char *input = malloc(1000*sizeof(char));

and if the string you are trying to copy to variable input exceeds the allocated memory size (strlen > 999: don't forget! String has a null terminator '\0' that takes up the additional 1 char space), just realloc as shown:

input = realloc(input, 1000*2*sizeof(char));
/* check if realloc works */
if (!input) {
    printf("Unexpected null pointer when realloc.\n");
    exit(EXIT_FAILURE);
}