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: 
`**` 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*`.
The statements made in your first code chunk are all correct.

```
char **char_ptrs = new char *[50];

```

...means you have an array of 50 `char *`s.

Your assesment of

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

```

is correct as well.

`**` simply means a 'pointer to a pointer'. It's not an operator.

When you write

```
new char *[50];

```

...you are saying 'allocate storage for 50 `char *`s'.