What is the best way to create a function object that includes additional functions as properties?

I am working with TypeScript and I want to design a module where a function is exported along with some additional functions added to it. Here is an example:

export default function log(msg: string) {
    console.log(msg);
}

//Property 'warn' does not exist on type '(msg: string) => void'.
log.warn = function(msg: string) {
    console.warn(msg);
};

To use this, you would do the following:

log('test');
log.warn('test');

How can I inform TypeScript that my function object includes extra properties so that it doesn't trigger the error message

Property 'warn' does not exist on type '(msg: string) => void'.
?

Answer №1

The correct way to write this would be:

function record(message: string) {

}
namespace record {
  export function caution(sign: string) { }
}

export default record;

Answer №2

Consider this format:

const record: { (data: string): void; details?: Function; } = function (data: string) {
  console.log(data);
}

record.details = function (data: string) { console.warn(data); };

export default record;

To put it differently, a function type with properties could be defined as

type FuncWithProp = {
  (FUNC_PARAMETERS): FUNC_RESPONSE_TYPE;
  YOUR_FIELD_HERE: SOME_KIND;
};

If you wish to specify the bound function more precisely to accept strings and produce them, while enabling additional ones, then

type Recorder = {
  (message: string): void;
  caution?: Recorder;
  problem?: Recorder;
};

const record: Recorder = function...
record.caution = function...
export default record;

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