在 Typescript 中传递 function 作为参数：预期为 0 arguments，但得到了 1.ts

Question

I am trying to pass in this doSomething function into performAction , but the error I'm getting is Expected 0 arguments, but got 1

type someType = {
  name: string,
  id: string
}

function doSomethingFn(props: someType) {
  console.log(props.name + " " + props.id);
}

function performAction(doSomething: () => void) {
  doSomething({
    name: "foo",
    id: "bar"
  });
}

performAction(doSomethingFn);

Am I using the proper syntax for Typescript?

Answer 1

The doSomething type seems incorrect. In the type declaration – () => void it takes no arguments, but later you are passing arguments to it.

For this piece of code following would work, but you would know better what should be the arguments and their types of doSomething . Probably use a interface if you already have an abstraction in your mind.

function performAction(doSomething: (stuff: { name: string, id: string }) => void) {
  doSomething({
    name: "foo",
    id: "bar"
  });
}

Demo for above.

Also if string in your code is a variable you need to change that because string is reserved for the type. In case you don't what to fix the type of doSomething right now you can use the Function type. Demo

Update

For your updated question you need to write function performAction(doSomething: (stuff: someType) => void Demo

Answer 2

You type doSomething as a function with no arguments: doSomething: () => void . Make it, dunno, doSomething: (arg: SomeInterface) => void (SomeInterface being, for example, {name: string; id: string;} ).

Answer 3

let a = 6;
let b = 2;

let res = outer(substr, a, b);
console.log(res);
res = outer(addit, a, b);
console.log(res);

function outer(inner: Function, a: number, b: number): number {
    return inner(a,b);
}

function substr(vala: number, valb: number): number {
    console.log('We are substracting');
    return vala - valb;
}

function addit(vala: number, valb: number): number {
    console.log('We are adding');
    return vala + valb;
}