How do you convert an unsigned int[16] of hexidecimal to an unsigned char array without losing any information?

I have a unsigned int[16] array that when printed out looks like this:

4418703544ED3F688AC208F53343AA59

The code used to print it out is this:

for (i = 0; i < 16; i++)
    printf("%X", CipherBlock[i] / 16), printf("%X",CipherBlock[i] % 16);
printf("\n");

I need to pass this unsigned int array "CipherBlock" into a decrypt() method that only takes unsigned char *. How do correctly memcpy everything from the "CipherBlock" array into an unsigned char array without losing information?

My understanding is an unsigned int is 4 bytes and unsigned char 1 byte. Since "CipherBlock" is 16 unsigned integers, the total size in bytes = 16 * 4 = 64 bytes. Does this mean my unsigned char[] array needs to be 64 in length?

If so, would the following work?

unsigned char arr[64] = { '\0' };
memcpy(arr,CipherBlock,64);

This does not seem to work. For some reason it only copies the the first byte of "CipherBlock" into "arr". The rest of "arr" is '\0' thereafter.

Solution

An int is at least 16 bits, same as a short in that regard.

It looks like every unsigned int has values 0-255 or 00-FF in your case, which is a safe range for an unsigned char. However, the proper way to convert one to the other is a cast:

for (int i=0; i<16; ++i) arr[i] = (unsigned char) CipherBlock[i];

But you have not specified what kind of data decrypt() expects. From the signature, I suspect integral data (strings are usually char* or const char*) but it's hard to be sure without a context.

Note that you could also do printf("%02X", CipherBlock[i]); for printing.