Specify a function's type inline with typescript

Question

Is there a way in typescript to define the type of a function (note: not an arrow function) in typescript?

I'm aware of the following method:

const MyFunctionCreator = (): MyFunction => {
    return function(input) {
        return "";
    };
};

However, I am trying to set the type of a static function inside a class, so this is not ideal.

class MyClass {
    static function(input) {
        return "";
    }
}

Is there a way in the example above to do something like:

class MyClass {
    static myFunction: MyFunction(input) {
        return "";
    }

    static myFunction(input) {
        return "";
    } as MyFunction
}

I can of course re-type the param/return types every time, but I wish to share types across my classes.

Not sure what you are looking for here ? Why can't you just define the function signature when you declare it ? you can reuse the type using a type query. — Titian Cernicova-Dragomir
– Titian Cernicova-Dragomir, Commented Jul 17, 2018 at 10:27
I want to assign a type to the function as a whole, I do not want to re-define the signature every time I create an identical function elsewhere. — ed'
– ed', Commented Jul 17, 2018 at 11:45

Titian Cernicova-Dragomir · Accepted Answer · 2018-07-17 13:30:06Z

1

You can use a function field instead of a member. For static fields there is not much difference, for instance fields they get assigned every time you create an object so that might have performance implications of you create a lot of instances.

type MyFunction = (input : string) => number

class MyClass {
    static myFunction: MyFunction = function (input) {
        return input.length; // input is string
    }

    // error wrong return type
    static myFunctionError: MyFunction = function (input) {
        return input; 
    }
}

answered Jul 17, 2018 at 13:30

Titian Cernicova-Dragomir

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2 Comments

ed' Over a year ago

I know about arrow functions, what I want to know is if you can do it without an arrow function

Titian Cernicova-Dragomir Over a year ago

It's not an arrow function, it's a function expression. But this is the only available option, 95% sure on this :)

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