Head First C: Erratum? Pointer Syntax

Question

I am teaching myself C programming, using "Head First C", published by O'Reilly. (And a couple of other texts, besides.)

I got very hung-up during the book's introduction to pointers, by the example program on p.58. What threw me was a little call-out in the text, pointing to the line:

int *choice = contestants;

The call-out reads:

“choice” is now the address of the “contestants” array.

And, as far as I can tell, that's wrong. That line assigns *choice to the value stored as the contestants array, is that not so?

Answer 1

The book is correct. Pointers work differently from variables, and * works differently in different circumstances.

To start, you probably know that a pointer is just a variable containing the address to the location in memory of another item

When we use int *choice we create a pointer to an integer, which stores the address of the integer

In this case, I'm assuming contestants is an array. We'll assume it's an array of ints

int contestants[5] = [0,1,2,3,4];

Collections, such as arrays, are always stored as a pointer to the first element. The second element is therefore addressof_first_item + sizeof(element_stored).

Let's assume the address is 0x12345, it's like this int contestants = 0x12345

And now we store that address in choice: int *choice = contestants

In other words, it's a pointer to an integer, which happens to be the first element in the contestants array.

As a side note, we could also write it as:

int* choice = &contestants[0];

The other use case for the * operator with regards to pointers, is dereferencing. Dereferencing involves getting the value at the address pointed to by a pointer.

int value = *choice;

will give us an integer with a value of 0, because the address stored points to the first element