Typescript interface default values

Question

I have the following interface in TypeScript:

interface IX {
    a: string,
    b: any,
    c: AnotherType
}

I declare a variable of that t

Answer 1

I stumbled on this while looking for a better way than what I had arrived at. Having read the answers and trying them out I thought it was worth posting what I was doing as the other answers didn't feel as succinct for me. It was important for me to only have to write a short amount of code each time I set up a new interface. I settled on...

Using a custom generic deepCopy function:

deepCopy = (input: any): T => {
  return JSON.parse(JSON.stringify(input));
};

Define your interface

interface IX {
    a: string;
    b: any;
    c: AnotherType;
}

... and define the defaults in a separate const.

const XDef : IX = {
    a: '',
    b: null,
    c: null,
};

Then init like this:

let x : IX = deepCopy(XDef);

That's all that's needed..

.. however ..

If you want to custom initialise any root element you can modify the deepCopy function to accept custom default values. The function becomes:

deepCopyAssign = (input: any, rootOverwrites?: any): T => {
  return JSON.parse(JSON.stringify({ ...input, ...rootOverwrites }));
};

Which can then be called like this instead:

let x : IX = deepCopyAssign(XDef, { a:'customInitValue' } );

Any other preferred way of deep copy would work. If only a shallow copy is needed then Object.assign would suffice, forgoing the need for the utility deepCopy or deepCopyAssign function.

let x : IX = object.assign({}, XDef, { a:'customInitValue' });

Known Issues

• It will not deep assign in this guise but it's not too difficult to modify deepCopyAssign to iterate and check types before assigning.
• Functions and references will be lost by the parse/stringify process. I don't need those for my task and neither did the OP.
• Custom init values are not hinted by the IDE or type checked when executed.