Accessing a C char array from a pointer

I am trying to pass a C char array to a function via its pointer and then accessing the char array in the function. At the moment I can only seem to access the first character and the rest is just random bytes. I have been trying to check the variables before and after the function call and they are all correct before the function call.

I think this question has the answer since it refers to dereferencing and only retaining the first character but when I try and implement the solution I am still getting a single character and not the whole string.

This is the code I have:

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

void hex_to_bytes(char *hex, char **bytes, size_t *bytes_len) {
    int length = strlen(hex) % 2 ? strlen(hex) + 1 : strlen(hex);
    unsigned char checkedHex[length];
    
    for (int i = 0; i <= length; i++) {
        if (i == 0 && strlen(hex) % 2){
            checkedHex[0] = '0';
        }
        else if (i == length){
            checkedHex[i] = '\x00';
        }
        else {
            checkedHex[i] = hex[i]; 
        } 
    }
    
    *bytes_len = length / 2;
    int counter = 0;
    for (int i = 0; i < length / 2; i++){
        sscanf(&checkedHex[counter], "%2hhx", &bytes[i]);
        counter += 2;
    }
}

void *b64_encode(unsigned char *bytes, size_t bytes_len)
{
    printf("in b64_encode %p\n", bytes);
    printf("bytes = %s\n", &bytes[48]);

    printf("0x");
    for (int i = 0; i < bytes_len; i++) {
        printf("%02X", bytes[i]);
    }
    printf("\n");
}

int main() {
    unsigned char hex[] = "49276d206b696c6c696e6720796f757220627261696e206c696b65206120706f69736f6e6f7573206d757368726f6f6d";
    unsigned char *bytes[48] = {0};
    size_t bytes_len = 0;

    hex_to_bytes(hex, (unsigned char **) &bytes, &bytes_len);

    printf("in main       %p\n", &bytes);
    printf("bytes = %s\n", &bytes[48]);

    unsigned char *base64 = b64_encode(bytes, bytes_len);
    return 1;
}

The output I get is:

in main       0x7ff7b0ef1350
bytes = 49276d206b696c6c696e6720796f757220627261696e206c696b65206120706f69736f6e6f7573206d757368726f6f6d
in b64_encode 0x7ff7b0ef1350
bytes = l
0x490000000000000027000000000000006D0000000000000020000000000000006B000000000000006900000000000000

>Solution :

unsigned char *bytes[48] is an array of pointers, not an array of chars. Get rid of the *.

And, in hex_to_bytes(), change the bytes parameter from char** to char*. You don’t need a pointer to a pointer.

Try this:

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

void hex_to_bytes(char *hex, char *bytes, size_t *bytes_len) {
    int length = strlen(hex);
    if (length % 2) ++length;
    unsigned char checkedHex[length];
    
    for (int i = 0; i <= length; i++) {
        if (i == 0 && strlen(hex) % 2){
            checkedHex[0] = '0';
        }
        else if (i == length){
            checkedHex[i] = '\x00';
        }
        else {
            checkedHex[i] = hex[i]; 
        } 
    }
    
    *bytes_len = length / 2;
    int counter = 0;
    for (int i = 0; i < length / 2; i++){
        sscanf(&checkedHex[counter], "%2hhx", &bytes[i]);
        counter += 2;
    }
}

void b64_encode(unsigned char *bytes, size_t bytes_len)
{
    printf("in b64_encode %p\n", bytes);
    printf("bytes = %.*s\n", bytes_len, bytes);

    printf("0x");
    for (int i = 0; i < bytes_len; i++) {
        printf("%02X", bytes[i]);
    }
    printf("\n");
}

int main() {
    unsigned char hex[] = "49276d206b696c6c696e6720796f757220627261696e206c696b65206120706f69736f6e6f7573206d757368726f6f6d";
    unsigned char bytes[48] = {0};
    size_t bytes_len = 0;

    hex_to_bytes(hex, bytes, &bytes_len);

    printf("in main       %p\n", bytes);
    printf("bytes = %.*s\n", bytes_len, bytes);

    b64_encode(bytes, bytes_len);
    return 1;
}