Allocating string with malloc

Question

I'm new in programming in C and now I'm studying strings. My question is: if I allocate a string using malloc (as in the code below), is the NULL character automatically inserted at the end of the string? I find an answer in another question here, and it seems that the NULL character is not automatically included. But here comes the problem: I know functions like strlen don't work if there isn't the NULL character, and in this code I use it and it works. So I think there is \0 at the end of my string, even if I don't write it anywhere. What's the answer?

Here's the code:

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

int main(int argc, char** argv) {
    char *stringa1;
    int n;
    int i;

    printf("How many characters in the string? ");
    scanf("%d", &n);

    stringa1 = (char*) malloc(n*sizeof(char));

    printf("Insert the string: ");
    scanf("%s", stringa1);

    free(stringa1);

    return 0;
}

Answer 1

malloc returns pointer to an uninitialized memory extent.

If you want that the memory extent would be initialized by zeroes then you can use another standard function calloc instead of malloc.

Take into account that usually such a question like this

printf("How many characters in the string? ");

imply that the terminating zero is not counted. So you have to allocate one more byte of memory. For example

stringa1 = ( char* )malloc( ( n + 1 ) *sizeof( char ) );

or

stringa1 = ( char* )calloc( n + 1, sizeof( char ) );

In the last case you may apply the function strlen which returns 0 because the memory extent is zero-initialized.

This call of scanf

scanf("%s", stringa1);

is unsafe. It is better to use fgets instead. For example

fgets( stringa1, n + 1, stdin );

This function can append the string with the new line character. To remove it from the string you can write

stringa1[strcspn( stringa1, "\n" )] = '\0';

Answer 2

malloc() returns a void* pointer to a block of memory stored in the heap. Allocating with malloc() does not initialize any string, only space waiting to be occupied.To add a null-terminating character, you either have to do this yourself, or use a function like scanf(), which adds this character for you. Having said this, you need to allocate space for this \0 character beforehand.

Your malloc() call should be this instead:

stringa1 = (char*) malloc((n+1)*sizeof(char)); /*+1 for '\0' character */

Note: You don't need to cast return of malloc. For more information, read this.

Another thing to point out is sizeof(char) is 1, so multiplying this in your malloc() call is not necessary.

You also need to check if malloc() returns NULL. This can be done like this:

if (stringa1 == NULL) {
    /* handle exit */

Also, you can only use strlen() on a null-terminated string, otherwise this ends up being undefined behaviour.

Once scanf() is called, and the stringa1 contains some characters, you can call strlen() on it.

Additionally, checking return of scanf() is also a good idea. You can check it like this:

if (scanf("%d", &n) != 1) {
    /* handle exit */

Your code with these changes:

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

int main(void) {
    char *stringa1 = NULL;
    size_t n, slen;

    printf("How many characters in the string? ");
    if (scanf("%zu", &n) != 1) {
        printf("Invalid input\n");
        exit(EXIT_FAILURE);
    }

    stringa1 = malloc(n+1);
    if (stringa1 == NULL) {
        printf("Cannot allocate %zu bytes for string\n", n+1);
        exit(EXIT_FAILURE);
    }

    printf("Insert the string: ");
    scanf("%s", stringa1);

    slen = strlen(stringa1);
    printf("String: %s Length: %zu\n", stringa1, slen);

    free(stringa1);
    stringa1 = NULL;

    return 0;
}