我可以使用扩展语法使用 TypeScript 制作包装函数，而不必指定参数和类型吗？

Question

The following code is a common, and very neat, way of implementing wrapper functions in JavaScript.

The code wraps innerFunction (which has some named arguments) with the wrapper outerFunction :

function innerFunction(firstArgument, secondArgument, thirdArgument) {
  console.log('innerFunction', arguments);
}

function outerFunction() {
  console.log('outerFunction', arguments);
  innerFunction(...arguments)
}

outerFunction(1, 2, 3);

This works perfectly fine as JavaScript - you can see outerFunction passes whatever arguments to innerFunction :

outerFunction [Arguments] { '0': 1, '1': 2, '2': 3 }
innerFunction [Arguments] { '0': 1, '1': 2, '2': 3 }

Typescript doesn't like this, as it wants me to put the inner functions types into the outer function.

Is there a better way of doing this in TypeScript? Surely TypeScript's static analysis can see the outer function gets its types from the inner function?

I accept the answer might be 'no, you have to add the types of the inner function to the outer function'. But I'd like to consult with my peers here in case there's a better way of doing this.

Answer 1

Typescript offers a utility type Parameters that gives you the types of the parameters of a function as a tuple. So you would type the outer function like the following:

function innerFunction(firstArgument: string, secondArgument: number, thirdArgument: boolean) {
  console.log('innerFunction', firstArgument, secondArgument, thirdArgument);
}

function outerFunction(...args: Parameters<typeof innerFunction>) {
  console.log('outerFunction', ...args);
  innerFunction(...args);
}