Typescript Is it possible to distinct generic argument void?

September 19, 2023

A function has generic type argument.

When context is void, I want to input 0 arguments, otherwise I want to input 1 argument.

If I used as context: Context | void in function argument, Even when I need to add context, But I can add void type.

is there any way to input 0 argument or check whether it is typed as void?

class TestClass<Context = void> {
  protected context : Context

  constructor(context: Context) {
    this.context = context;
  }
}

export function genericVoidNeedArgument<Context = void>(
  context: Context,
) {
  // can check type..?
  return new TestClass(context);
}

type UserType = {
  id: string
  // and so on..
};

// expected : o, real : error
// An argument for 'context' was not provided.
genericVoidNeedArgument();

// expected : error, real : error
genericVoidNeedArgument<UserType>();

// expected : o, real : o
genericVoidNeedArgument<UserType>({id: "123"});

>Solution :

You can use Function Overloads to do that.

class TestClass<Context = void> {
  protected context: Context;

  constructor(context: Context) {
    this.context = context;
  }
}

function genericVoidNeedArgument(): TestClass<void>;

function genericVoidNeedArgument<Context>(
  context: Context
): TestClass<Context>;

function genericVoidNeedArgument(context?: any): any {
  return new TestClass(context);
}

type UserType = {
  id: string;
};

genericVoidNeedArgument();
genericVoidNeedArgument<UserType>({id: "123"});
genericVoidNeedArgument<UserType>();
// ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ -> Expected 1 arguments, but got 0.

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