Question

char firstName[32];

I understand that each char occupies 1 byte in memory. So does the above occupy 32 bytes of memory?

Am I missing a pointer that takes up memory too or is this just 32 bytes?

Answer 1

No, that takes up exactly 32 bytes of memory.
There is no pointer.

This is often an area of confusion, since an array name silently “decays” to a “char*”

char* fname = firstName;

So, firstName may be of type const char*, but it is not itself a char* variable. It is exactly like:

 int x = 5;

x is int variable and takes up space. 5 on the other hand, is just a constant value of int type. It takes of no space; it’s just a value.