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Open
5 tasks done
tolotrasamuel opened this issue Jul 16, 2020 · 2 comments
Open
5 tasks done

Cast Method of a class to a certain type #39623

tolotrasamuel opened this issue Jul 16, 2020 · 2 comments
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Awaiting More Feedback This means we'd like to hear from more people who would be helped by this feature Suggestion An idea for TypeScript

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@tolotrasamuel
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tolotrasamuel commented Jul 16, 2020

Search Terms

casting
type
interface
method
function
parameters
inference
class

Suggestion

Currently, it is impossible to infer the parameters and the return types of a method of a class using a type

The goal is to be able to declare a method without repeating the param types and the return type. I used typescript Type for that.

But there is currently no way to do this without changing the compiled javascript code.

type IBar = (x:number, y: number)=>number;

I tried

type IBar = (x:number, y: number)=>number;
class Foo {
    sum:IBar=(x,y)=>{
        return x+y;
    }
}

BUT unfortunately, this changes the javascript code compiled into:

class Foo {
    constructor() {
        this.sum = (x, y) => {
            return x + y;
        };
    }
}

I am looking for something that compiles into:

class Foo {
    sum(x, y) {
        return x + y;
    }
}

Another option currently is to declare an interface and implement it on the class like so:

interface IBar { sum(x: number, y: number): number; }

class Foo implements IBar {
    public sum(x: number, y: number): number {
        return x + y;
    }
}

HOWEVER, this approach would bnd the method name to only sum . What if I want to use the same method type with a different name like divide or multiply

Use Cases

It will be used like this

class Foo {
    sum(x,y){
        return x+y;
    } as IBar
}

Examples

class Foo {
    sum(x,y){
        return x+y;
    } as IBar
}

or

class Foo {
    divide(x,y){
        return x+y;
    } as IBar
}

or

class Foo {
    multiply(x,y){
        return x+y;
    } as IBar
}

Checklist

My suggestion meets these guidelines:

  • This wouldn't change the runtime behavior of existing JavaScript code
  • This could be implemented without emitting different JS based on the types of the expressions
  • This isn't a runtime feature (e.g. library functionality, non-ECMAScript syntax with JavaScript output, etc.)
  • This feature would agree with the rest of TypeScript's Design Goals.
  • This wouldn't be a breaking change in existing TypeScript/JavaScript code
@Barbiero
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Barbiero commented Jul 17, 2020

What do we think of this form?

type BinaryOperator<T> = (x: T, y: T) => T

class Foo {
    // Notice the lack of "="
    sum: BinaryOperator<number> (x, y) {
      return x+y;
    }
}

This would be a nice parallel to usual variable type declaration. I feel like the as X expression is an assertion instead of a declaration. Like, if you were to use it on an arrow function:

const sum = ((x,y,z) => false) as BinaryOperator<number> // sum is (number,number) => number

const sum: BinaryOperator<number> = ((x,y,z) => false) //error!

I do like the idea of declaring the typing for methods though.

@RyanCavanaugh RyanCavanaugh added Awaiting More Feedback This means we'd like to hear from more people who would be helped by this feature Suggestion An idea for TypeScript labels Jul 17, 2020
@bodinsamuel
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+1 on this
It would be very useful

class Foo {
    // Notice the lack of "="
    sum: BinaryOperator<number> (x, y) {
      return x+y;
    }
}

This solution feels good to me 👍🏻


Note that I would also expect to work with (or as an alternative) implements

interface IFoo {
  sum: BinaryOperator<number>
}

class Foo implements IFoo {
  sum() (x, y) {
      return x+y;
    }
}

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