Question: I want to create a function object, which also has some properties held on it. For example in JavaScript I would do:

```
var f = function() { }
f.someValue = 3;

```

Now in TypeScript I can describe the type of this as:

```
var f: { (): any; someValue: number; };

```

However I can't actually build it, without requiring a cast. Such as:

```
var f: { (): any; someValue: number; } =
    <{ (): any; someValue: number; }>(
        function() { }
    );
f.someValue = 3;

```

How would you build this without a cast?

Answer: 
So if the requirement is to simply build and assign that function to "f" without a cast, here is a possible solution:

```
var f: { (): any; someValue: number; };

f = (() => {
    var _f : any = function () { };
    _f.someValue = 3;
    return _f;
})();

```

Essentially, it uses a self executing function literal to "construct" an object that will match that signature before the assignment is done. The only weirdness is that the inner declaration of the function needs to be of type 'any', otherwise the compiler cries that you're assigning to a property which does not exist on the object yet.

EDIT: Simplified the code a bit.
As a shortcut, you can dynamically assign the object value using the ['property'] accessor:

```
var f = function() { }
f['someValue'] = 3;

```

This bypasses the type checking. However, it is pretty safe because you have to intentionally access the property the same way:

```
var val = f.someValue; // This won't work
var val = f['someValue']; // Yeah, I meant to do that

```

However, if you really want the type checking for the property value, this won't work.