Understanding C++ pointers (when they point to a pointer)

Question

I think I understand references and pointers pretty well. Here is what I (think I) know:

int i = 5; //i is a primitive type, the value is 5, i do not know the address.
int *ptr;  //a pointer to an int. i have no way if knowing the value yet.
ptr = &i;  //now i have an address for the value of i (called ptr)
*ptr = 10; //Go to the value in ptr, use it to find a location and store 10 there

Please feel free to comment or correct these statements.

Now I'm trying to make the jump to arrays of pointers. Here is what I do not know:

char **char_ptrs = new char *[50];
Node **node_ptrs = new Node *[50];

My understanding is that I have 2 arrays of pointers, one set of pointers to chars and one to nodes. So if I wanted to set the values, I would do something like this:

char_ptrs[0] = new char[20];
node_ptrs[0] = new Node;

Now I have a pointer, in the 0 position of my array, in each respective array. Again, feel free to comment here if I'm confused.

So, what does the ** operator do? Likewise, what is putting a single * next to the instantiation doing (*[50])? (what is that called exactly, instantiation?)

Answer 1

A few comments:

*ptr = 10; // Doesn't need to "go get" the value. Just overwrites it.

Also:

char **char_ptrs = new char *[50];
Node **node_ptrs = new Node *[50];

It is easier to think that you have two arrays. However, technically (and as far as the compiler is concerned) what you have is two pointers. One is a pointer to a (pointer to a char) and the other is a pointer to a (pointer to a node).

This is easily seen by the declarations of your variables, which, by the way, can be most easily read right-to-left:

char **char_ptrs

Reading right to left: char_ptrs is a pointer to a pointer to char

Putting a * next to a pointer is properly called dereferencing that pointer. Since arrays do not technically exist, operator [] on arrays is also a dereferencing operation: arr[i] is another way of writing *(arr + i). To properly understand this, you need to be familiar with pointer arithmetic.

More than one consecutive *s: each one dereferences the result of the expression it operates on. So when writing:

char c = **char_ptrs;

what happens is:

char_ptrs is a pointer to a pointer to a char. Dereferencing it once (for the rightmost *) gets you its value, which is a pointer to a char. Dereferencing that value (for the leftmost *) gives you its own value in turn, which is a char. In the end, c contains the value of the char stored in memory at the place where the pointer pointed to by char_ptrs (in other words, the first pointer in your array) points.

Conversely, if you write **char_ptrs = 'a'; then you are changing the value in that memory location.

Answer 2

** is just * twice, so a pointer to a pointer.

When put next to a type, * binds left, not right. Saying new char *[50] is actually new char* [50] and instantiates an array of 50 char*.