What does typedef int var[1]; do?

Question

I've encountered the following code:

typedef int var[1]; // or var[3]

what does it actually do? I don't understand what does the subscript add, since now I can define "var" for int

Answer 1

typedef declarations use the same syntax as ordinary variable declarations. The difference is that instead of declaring "a variable named x of type y," you declare "a type named x that is a synonym for type y." The syntax is otherwise the same.

So, let's remove the typedef from your example and see what we get:

int var[1];

var is a variable whose type is int[1], or, an array of one int. If we add the typedef back:

typedef int var[1];

this makes var a synonym for the type int[1].

The same works for other kinds of ugly or complex types:

int (*fp)(int);         // fp is a function pointer variable
typedef int (*fp)(int); // fp is a function pointer type

---

You can avoid most of this confusion by using an identity template, declared as

template <typename T> struct identity { typedef T type; };

Using this template, the meaning of a complex type or variable declaration is much clearer. For example, we can declare our types like so:

typedef identity<int[1]     >::type var; // array type
typedef identity<int(*)(int)>::type fp;  // function pointer type

and because variable declarations use the same syntax as type declarations, we can declare variables that have a complex using identity as well:

identity<int[1]     >::type var; // array variable
identity<int(*)(int)>::type fp;  // function pointer variable

Answer 2

var is a type definition for an int array of size one.

You can write

var x;

and x will be a variable of type int[1]