Alternate function overloading in Typescript?

Question

An undocumented feature of the language seems to be using the pipe operator to overload arguments.

Example:

 function foo(user : string | number) {
    //...
 }

Seems to be working fine until the case below. My question is (1) is it safe to continue to use the pipe operator this way? (2) if so, how can I fix the case below?

 function _isString<T>(value : T) : boolean { return typeof value === 'string'; };

 function foo(services : string | string[]) {

    //doesn't compile
    const aService : string[] = _isString(services) ? [services] : services;

    //but this does
    const bService : string[] = typeof services === 'string' ? [services] : services;
 }

basarat · Accepted Answer · 2015-12-10 04:44:20Z

1

My question is (1) is it safe to continue to use the pipe operator this way?

Yes. It is by design.

(2) if so, how can I fix the case below?

You need a user defined type guard function. Here is the fixed code:

 function _isString(value : any) : value is string { return typeof value === 'string'; };

 function foo(services : string | string[]) {

    // works
    const aService : string[] = _isString(services) ? [services] : services;

    // works
    const bService : string[] = typeof services === 'string' ? [services] : services;
 }

answered Dec 10, 2015 at 4:44

basarat

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1 Comment

U Avalos Over a year ago

Works like a charm. Did not know about TS guard functions... guess the docs are very outdated.

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