方法的泛型类型取决于 TypeScript 中 Class 的泛型类型

Question

I have a use case that can be expressed in pseudo-code as follows:

class Gen<T> {
  public doStuff<U>(input: U 
         /* If T is an instance of number, 
          then input type U should be an instance of custom type ABC, or
          If T is an instance of string,
          then input type U should an instance of custom type XYZ, else
          compile error */) {

     // do stuff with input
  }
}

Can this be expressed in TypeScript?

Answer 1

Absolutely, with inference this is easily feasible. TypeScript allows you to "return" a different type based on a generic input type checking. You should just use a InferInputType<T> type like this:

type InferInputType<T> =
    T extends number ? ABC :
    T extends string ? XYZ :
    never;

then you can rewrite your Gen as:

class Gen<T> {
    public doStuff(input: InferInputType<T>) {}
}

Then you can use your class like this:

const genNumber = new Gen<number>();
genNumber.doStuff({ value: 10 });
genNumber.doStuff({ value: 'abc' }); // Error

You can see a working example in the playground: Playground Link