What is the proper way to incorporate generics into a function in TypeScript when you plan to call it using .call()?

Question

What is the proper way to incorporate generics into a function in TypeScript when you plan to call it using .call()?

interface Wrapped<T> {
  data: T;
}

interface BetterWrapper<T> {
  betterData: T;
}

function abc<T>(test: Wrapped<T>): BetterWrapper<T> {
  return {betterData: test.data}
}

const result = abc<string>.apply({}, { data: 'hello' });

When I encounter the error message stating "cannot find name call" while working with the code above, I believe it may be due to the use of an incorrect method. In this case, since the return type is a string, the correct approach in TypeScript is to use the apply method instead of call.

typescript

Answer 1

Answer №1

To tackle this problem, you can define an interface that includes the necessary generic signature for the call method. This allows you to call the method either with explicit type parameters like abc.call<string>(...) or without them to let the type be inferred automatically.

interface WrapperInterface {
  <T>(input: Wrapped<T>): BetterWrapper<T>;
  call<T>(target: object, argument: Wrapped<T>): BetterWrapper<T>;
}
const abc: WrapperInterface = input => ({ betterData: input.data });

// outcome: BetterWrapper<string>
const outcome = abc.call({}, { data: 'world' });

Practice Link

Answer 2

To tackle this problem, you can define an interface that includes the necessary generic signature for the call method. This allows you to call the method either with explicit type parameters like abc.call<string>(...) or without them to let the type be inferred automatically.

interface WrapperInterface {
  <T>(input: Wrapped<T>): BetterWrapper<T>;
  call<T>(target: object, argument: Wrapped<T>): BetterWrapper<T>;
}
const abc: WrapperInterface = input => ({ betterData: input.data });

// outcome: BetterWrapper<string>
const outcome = abc.call({}, { data: 'world' });

Practice Link

What is the proper way to incorporate generics into a function in TypeScript when you plan to call it using .call()?

Answer №1

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