Why does NULL terminate the loss of the sha256 hash when using dynamic memory but not static memory?

Question

I use the libsrypto openssl library to try to create the sha256 hash of an arbitrary input buffer, however, when I null complete the hash buffer to make it a C string, it fails, but only when using the dynamic memory returned from mallocbut not when I use statically allocated memory, as in this example: generate sha256 using openssl and C ++ . I allocate 65 bytes, as in the example, as shown by the printed value length. So, why does zero terminate buffer failure when using dynamic memory as opposed to static memory?

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <openssl/sha.h>

static int sha256(char *in, char **out)
{
    int rtrn, i;
    SHA256_CTX sha256;
    unsigned char hash[SHA256_DIGEST_LENGTH];

    rtrn = SHA256_Init(&sha256);
    if(rtrn < 0)
    {
        printf("Sha Init Error\n");
        return -1;
    }

    rtrn = SHA256_Update(&sha256, in, strlen(in));
    if(rtrn < 0)
    {
        printf("Sha Update Error\n");
        return -1;
    }

    rtrn = SHA256_Final(hash, &sha256);
    if(rtrn < 0)
    {
        printf("Sha Final Error\n");
        return -1;
    }

    *out = malloc((SHA256_DIGEST_LENGTH * 2) + 1);
    if(*out == NULL)
    {
        printf("Can't allocate output buf\n");
        return -1;
    }

    for(i = 0; i < SHA256_DIGEST_LENGTH; i++)
    {
        sprintf(*out + (i * 2), "%02x", hash[i]);
    }

    printf("Length: %d\n", (SHA256_DIGEST_LENGTH * 2) + 1);

    *out[64] = '\0';

    return 0;
}

int main(void)
{
    int rtrn;
    char *input = "3r98h932hr934hor";
    char *output;

    rtrn = sha256(input, &output);
    if(rtrn < 0)
    {
        printf("Sha256\n");
        return -1;
    }

    printf("%s\n", output);

    return 0;
}

Answer 1

, , :

*out[64] = '\0';

:

(*out)[64] = '\0';

([]) , (*) C, , , :

*(out[64]) = '\0';

( char **).