Typescript function overloading with object

Question

I have a function that accepts an object as the only parameter. For example:

interface MyMap<T> {
  [id: string]: T;
}

type Options = {
  asObject: boolean,
  other?: Function
};

  function get(options: Options): any[];
  function get(options: Options): MyMap<any>;
  function get(options: Options): any[] | MyMap<any>;
  function get(options: Options = {asObject: false, other: () => {}}): any[] | MyMap<any> {
    if (options.asObject) return {} as MyMap<any>;

    return [];
}


const result = get({asObject: true});

When the asObject value is true, I want to infer the type to MyMap. I know how to do it with simple boolean value, but how can I accomplish this with an object?

Aluan Haddad · Accepted Answer · 2018-05-12 18:53:25Z

2

function get(options: {asObject: true}): MyMap<any>;
function get(options: {asObjectb false}): any[];
function get(): any[];
function get(options: Options = {asObject: false, other: () => {}}): any[] | MyMap<any> {
   //  ... 
}

Should be all that you need. This will overload the function for the literal types true and false of the property asObject.

Note that when overloads are declared, the implementation itself does not contribute a signature to the overload set.

edited May 12, 2018 at 18:53

answered May 12, 2018 at 18:32

Aluan Haddad

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10 Comments

undefined Over a year ago

The function can return an array. It depends on the asObject property.

Aluan Haddad Over a year ago

No problem. Glad to help

undefined Over a year ago

But why it's not recognize the default value? function get(options: Options = {asObject: false, other: () => {}}): any[] | MyMap<any> { }

undefined Over a year ago

This still infered as MyMap

Aluan Haddad Over a year ago

Because the implementation signature of an overload does not contribute to the overload set.

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