Typescript – Are strongly-typed functions as parameters possible in TypeScript

typescript

In TypeScript, I can declare a parameter of a function as a type Function. Is there a "type-safe" way of doing this that I am missing? For example, consider this:

class Foo {
    save(callback: Function) : void {
        //Do the save
        var result : number = 42; //We get a number from the save operation
        //Can I at compile-time ensure the callback accepts a single parameter of type number somehow?
        callback(result);
    }
}

var foo = new Foo();
var callback = (result: string) : void => {
    alert(result);
}
foo.save(callback);

The save callback is not type safe, I am giving it a callback function where the function's parameter is a string but I am passing it a number, and compiles with no errors. Can I make the result parameter in save a type-safe function?

TL;DR version: is there an equivalent of a .NET delegate in TypeScript?

Best Answer

Sure. A function's type consists of the types of its argument and its return type. Here we specify that the callback parameter's type must be "function that accepts a number and returns type any":

class Foo {
    save(callback: (n: number) => any) : void {
        callback(42);
    }
}
var foo = new Foo();

var strCallback = (result: string) : void => {
    alert(result);
}
var numCallback = (result: number) : void => {
    alert(result.toString());
}

foo.save(strCallback); // not OK
foo.save(numCallback); // OK

If you want, you can define a type alias to encapsulate this:

type NumberCallback = (n: number) => any;

class Foo {
    // Equivalent
    save(callback: NumberCallback) : void {
        callback(42);
    }
}
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