I'm trying to get the type of an instance method of a class. Is there a built-in (better) way other than looking up the type in the prototype of the class?
class MyClass
{
private delegate: typeof MyClass.prototype.myMethod; // gets the type ( boolean ) => number;
public myMethod( arg: boolean )
{
return 3.14;
}
}
Thanks in advance!
You could use TypeScript's built in InstanceType for that:
class MyClass
{
private delegate: InstanceType<typeof MyClass>['myMethod']; // gets the type (boolean) => number;
public myMethod( arg: boolean )
{
return 3.14;
}
}
MyClass has generic type MyClass<T>. It's easier to just use MyClass<T>['method']A simple, idiomatic approach appears to be:
type MyMethodType = MyClass['myMethod'];
It doesn't seem to have any downsides, and, as @cuddlefish points out, it also works if MyClass is generic.
you can use the following type:
type TypeOfClassMethod<T, M extends keyof T> = T[M] extends Function ? T[M] : never;
With that, you can write the following:
class MyClass
{
private delegate: TypeOfClassMethod<MyClass, 'myMethod'>; // gets the type (boolean) => number;
public myMethod( arg: boolean )
{
return 3.14;
}
}
MyClass["myMethod"]? Is there any obvious drawback doing it this way as opposed to the one in the answer?
typeof MyClass.prototype.myMethod. I don't see drawbacks - just a different way to achieve this same thing.If you want to have a private method but still being able to pull this trick and have it have exposed as public then you can do this:
class MyClass {
public myMethodType: typeof MyClass.prototype.myMethod;
private myMethod(arg: boolean) {
return 3.14;
}
}
let fn: typeof MyClass.prototype.myMethodType;
That compiles to:
var MyClass = (function () {
function MyClass() {
}
MyClass.prototype.myMethod = function (arg) {
return 3.14;
};
return MyClass;
}());
var fn;
As you can see, the myMethodType member isn't part of the compiled js, which is good because it's only used for its type.
