Build a function object with properties in TypeScr

问题:

I want to create a function object, which also has some properties held on it. For example in JavaScript I would do:

var f = function() { }
f.someValue = 3;

Now in TypeScript I can describe the type of this as:

var f: { (): any; someValue: number; };

However I can't actually build it, without requiring a cast. Such as:

var f: { (): any; someValue: number; } =
    <{ (): any; someValue: number; }>(
        function() { }
    );
f.someValue = 3;

How would you build this without a cast?

回答1:

So if the requirement is to simply build and assign that function to "f" without a cast, here is a possible solution:

var f: { (): any; someValue: number; };

f = (() => {
    var _f : any = function () { };
    _f.someValue = 3;
    return _f;
})();

Essentially, it uses a self executing function literal to "construct" an object that will match that signature before the assignment is done. The only weirdness is that the inner declaration of the function needs to be of type 'any', otherwise the compiler cries that you're assigning to a property which does not exist on the object yet.

EDIT: Simplified the code a bit.

回答2:

Update: This answer was the best solution in earlier versions of TypeScript, but there are better options available in newer versions (see other answers).

The accepted answer works and might be required in some situations, but have the downside of providing no type safety for building up the object. This technique will at least throw a type error if you attempt to add an undefined property.

interface F { (): any; someValue: number; }

var f = <F>function () { }
f.someValue = 3

// type error
f.notDeclard = 3

回答3:

TypeScript is designed to handle this case through declaration merging:

you may also be familiar with JavaScript practice of creating a function and then extending the function further by adding properties onto the function. TypeScript uses declaration merging to build up definitions like this in a type-safe way.

Declaration merging lets us say that something is both a function and a namespace (internal module):

function f() { }
namespace f {
    export var someValue = 3;
}

This preserves typing and lets us write both f() and f.someValue. When writing a .d.ts file for existing JavaScript code, use declare:

declare function f(): void;
declare namespace f {
    export var someValue: number;
}

Adding properties to functions is often a confusing or unexpected pattern in TypeScript, so try to avoid it, but it can be necessary when using or converting older JS code. This is one of the only times it would be appropriate to mix internal modules (namespaces) with external.

回答4:

This is easily achievable now (typescript 2.x) with Object.assign(target, source)

example:

The magic here is that Object.assign<T, U>(t: T, u: U) is typed to return the intersection T & U.

Enforcing that this resolves to a known interface is also straight-forward. For example:

interface Foo {
  (a: number, b: string): string[];
  foo: string;
}

let method: Foo = Object.assign(
  (a: number, b: string) => { return a * a; },
  { foo: 10 }
); 

which errors due to incompatible typing:

Error: foo:number not assignable to foo:string
Error: number not assignable to string[] (return type)

caveat: you may need to polyfill Object.assign if targeting older browsers.

回答5:

As a shortcut, you can dynamically assign the object value using the ['property'] accessor:

var f = function() { }
f['someValue'] = 3;

This bypasses the type checking. However, it is pretty safe because you have to intentionally access the property the same way:

var val = f.someValue; // This won't work
var val = f['someValue']; // Yeah, I meant to do that

However, if you really want the type checking for the property value, this won't work.

回答6:

An updated answer: since the addition of intersection types via &, it is possible to "merge" two inferred types on the fly.

Here's a general helper that reads the properties of some object from and copies them over an object onto. It returns the same object onto but with a new type that includes both sets of properties, so correctly describing the runtime behaviour:

function merge<T1, T2>(onto: T1, from: T2): T1 & T2 {
    Object.keys(from).forEach(key => onto[key] = from[key]);
    return onto as T1 & T2;
}

This low-level helper does still perform a type-assertion, but it is type-safe by design. With this helper in place, we have an operator that we can use to solve the OP's problem with full type safety:

interface Foo {
    (message: string): void;
    bar(count: number): void;
}

const foo: Foo = merge(
    (message: string) => console.log(`message is ${message}`), {
        bar(count: number) {
            console.log(`bar was passed ${count}`)
        }
    }
);

Click here to try it out in the TypeScript Playground. Note that we have constrained foo to be of type Foo, so the result of merge has to be a complete Foo. So if you rename bar to bad then you get a type error.

NB There is still one type hole here, however. TypeScript doesn't provide a way to constrain a type parameter to be "not a function". So you could get confused and pass your function as the second argument to merge, and that wouldn't work. So until this can be declared, we have to catch it at runtime:

function merge<T1, T2>(onto: T1, from: T2): T1 & T2 {
    if (typeof from !== "object" || from instanceof Array) {
        throw new Error("merge: 'from' must be an ordinary object");
    }
    Object.keys(from).forEach(key => onto[key] = from[key]);
    return onto as T1 & T2;
}

回答7:

I can't say that it's very straightforward but it's definitely possible:

interface Optional {
  <T>(value?: T): OptionalMonad<T>;
  empty(): OptionalMonad<any>;
}

const Optional = (<T>(value?: T) => OptionalCreator(value)) as Optional;
Optional.empty = () => OptionalCreator();

if you got curious this is from a gist of mine with the TypeScript/JavaScript version of Optional

回答8:

This departs from strong typing, but you can do

var f: any = function() { }
f.someValue = 3;

if you are trying to get around oppressive strong typing like I was when I found this question. Sadly this is a case TypeScript fails on perfectly valid JavaScript so you have to you tell TypeScript to back off.

"You JavaScript is perfectly valid TypeScript" evaluates to false. (Note: using 0.95)

回答9:

Old question, but for versions of TypeScript starting with 3.1, you can simply do the property assignment as you would in plain JS, as long as you use a function declaration or the const keyword for your variable:

function f () {}
f.someValue = 3; // fine
const g = function () {};
g.someValue = 3; // also fine
var h = function () {};
h.someValue = 3; // Error: "Property 'someValue' does not exist on type '() => void'"

Reference and online example.