How do I override a property in typescript?

This may not be possible due to current language limitations, but I'm using the latest TS (1.8.10) and am running into an issue with the ui-grid typings. The isRowSelectable property on IGridOptions is defined as an optional boolean but the documentation says it's a function (and it is). I am trying to override the boolean property to be a function that returns a boolean.

Normally, I just extend the typing interface and do what I need, but that's not working in this case. Here's what I have:

interface ISelectionGridOptions extends IGridOptions {
  isRowSelectable: (row: IGridRowOf<Profile>) => boolean;
}

Where the relevant field in IGridOptions is:

export interface IGridOptions {
  ...
  isRowSelectable?: boolean
  ...
}

And the error I'm getting is:

(41,11): error TS2430: Interface 'ISelectionGridOptions' incorrectly extends interface 'IGridOptionsOf<any>'.
  Types of property 'isRowSelectable' are incompatible.
    Type '(row: IGridRowOf<Profile>) => boolean' is not assignable to type 'boolean'.

Short of fixing the core typings definitions, is there a way to fix this in my code?

Solution

If the type definitions are incorrect, you can't use overriding in order to fix them - the type system correctly treats this as an error. In a typed language, it's a good thing that subclasses are prevented from changing type signatures. The problem here is the broken type definition. If it's not practical to fix the def, and the issue occurs in only a few places in your code, you could use a cast to any to disable typechecking until the defs are fixed:

var foo = <boolean> (<any>bar).isRowSelectable(args);
bar.isRowSelectable = <any>foo;

Just leave a comment explaining what's going on. It is, of course, best to fix the type defs.

Note: It breaks, like the error says, because a function isn't assignable to boolean. The function peg is failing to fit into the boolean hole, not vice versa