Dynamically allocate an array of strings

Question

How can I fix this code in a way that it prints the words in the array? Moreover this is the right way to dynamically allocate memory for n words of max size 40?

int main() {

    int n;
    char *arr;
    int i;

    printf("Give me a number:");
    scanf("%d", &n);

    arr = malloc(n * 40);

    for (i = 0; i < n; i++)
    {
        printf("Give me a word: ");
        scanf("%s", &arr[i]);
    }

    for (i = 0; i < n; i++)
    {
        printf("%s", arr[i]); //< --problem here
    }

    return 0;
}

Answer 1

Your allocation is not the best, and printf argument arr[i] expects a char* but you pass it an int (a char if you'd like).

Here is how you should do it, with comments:

Live demo

#include <stdio.h>
#include <stdlib.h> //for malloc

int main(){
    int n;
    int i;
    
    printf("Give me a number:");
    scanf("%d", &n);

    //declare a variable of pointer to pointer to char
    //allocate memory for the array of pointers to char, 
    //each one capable of pointing to a char array
    char **arr = malloc(n * sizeof *arr);

    if(arr == NULL){ //check for allocation errors
        perror("malloc");
        return EXIT_FAILURE;
    }

    //allocate memory for each individual char array
    for(i = 0; i < n; i++){
        arr[i] = malloc(40); //char size is always 1 byte

        if(arr == NULL){ //check for allocation errors
            perror("malloc");
            return EXIT_FAILURE;
        }
    }

    for (i = 0; i < n; i++){
        printf("Give me a word: ");
        //limit the size of read input to 39 charaters to avoid overflow, 
        //a nul character will be added by scanf
        scanf("%39s", arr[i]);
    }

    for (i = 0; i < n; i++){
        printf("%s\n", arr[i]);
       
    }
     
    for(int i = 0; i < n; i++){ //free the memory for each char array
        free(arr[i]);
    }
    free(arr); //free array of pointers

    return 0;
}

You can also do this with less code using a pointer to array of 40 chars, this will simplify the memory allocation and deallocation:

Sample with comments:

Live demo

#include <stdio.h>
#include <stdlib.h> //for malloc

int main(){
    int n;    
    int i;

    printf("Give me a number:");
    scanf("%d", &n);

    //declare a pointer to array of chars and
    //allocate memory for all the char arrays
    char (*arr)[40] = malloc(n * sizeof *arr);

    if(arr == NULL){ //check for allocation errors
        perror("malloc");
        return EXIT_FAILURE;
    }

    for (i = 0; i < n; i++){
        printf("Give me a word: ");
        scanf("%39s", arr[i]);
    }

    for (i = 0; i < n; i++){
        printf("%s\n", arr[i]);
    }

    free(arr); //free allocated memory

    return 0;
}

Answer 2

This:

for(i=0;i<n;i++){
    printf("Give me a word: ");
    scanf("%s",&arr[i]);
}

is probably not what you want.
You probably want this instead:

for(i=0; i<n; i++){
    printf("Give me a word: ");
    scanf("%s", arr + i*40);
}

then later:

for(i=0; i<n; i++){
    printf("%s", arr + i*40);
}

Remember that a string in C is just an array of characters. Thus when defining char *arr, you are creating a single string. Had you done char **arr it would have been an array of strings, which is what you want.
However, I find that allocating/freeing arrays of arrays on the heap to be rather inconvenient, and prefer 'flattening' them into a single array.
This is exactly what you were doing with arr = malloc(n*40), but then later you treated this array of characters as an array of strings when you did this:

for(i=0; i<n; i++){
    printf("Give me a word: ");
    scanf("%s", &arr[i]);
}

Note that this is actually perfectly legal (scanf wanted a char* and you gave it one), but it's a logical error since you are giving it the n-th character when you wanted to give it the n-th array.

And oh yes, don't forget to free(arr) later.