char array not displaying when passed to a function - c

Question

Why is the passing of char array not showing?
The location of the pointer is passed to a function.
char plaintext[] = {
0xCD, 0x76, 0x43, 0xF0,
0x72, 0xA4, 0xA0, 0x82,
}
Given
void displayArray(char** plaintext, int size) {
// int newSize = sizeof(**plaintext);
int i;
for(i=0; i < size; ++i) {
printf("%02X ", (0xff & *plaintext[i]));
if (((i+1)% 8) == 0) // as index starts from 0, (i+1)
printf("\n");
}
}
in main()
char* plaintextCopy;
plaintextCopy = (char*) malloc(numberOfItems*sizeof(char));
memcpy(plaintextCopy, plaintext, numberOfItems);
displayArray(&plaintextCopy, numberOfItems);

Answer 1

Based on your code :
void displayArray(char** plaintext, int size)
{
int i;
for(i=0; i < size; i++)
{
printf("%02X ", (0xff & (*plaintext)[i]));
if(((i+1)% 8) == 0) // as index starts from 0, (i+1)
printf("\n");
}
}
int main(void)
{
char plaintext[] = {
0xCD, 0x76, 0x43, 0xF0,
0x72, 0xA4, 0xA0, 0x82,
};
int numberOfItems = sizeof(plaintext);
char* plaintextCopy;
plaintextCopy = (char*) malloc(numberOfItems*sizeof(char));
memcpy(plaintextCopy, plaintext, numberOfItems);
displayArray(&plaintextCopy, numberOfItems);
return 0;
}
It outputs :
CD 76 43 F0 72 A4 A0 82
Also, if you're sending an array that you want to display or change the values of, you don't need to send a double pointer to a function. A regular pointer would do. You should only use double pointers if the original array changes it's location in memory, that is, it's getting a new pointer after the function returns.