typescript 将 function 标记为实现接口

Question

I want to indicate that a function is expected to implement a certain interface.

For example:

import { BrowserEvents, eventHandler, Event } from './browser-events'; 

export function setup(){
 const browserEvents = new BrowserEvents();
 browserEvents.onClicks(handleClicks);

 function handleClicks(ev: Event) {
   ev.preventDefault();
 }
}

I would like to tell typescript that handleClicks should match type eventHandler . Obviously, if the signature of handleClicks doesn't match onClicks , ts will throw an error. But I would like to show the error 'at the source' like this syntax:

// (not valid syntax!)
function handleClicks(ev: Event) implements eventHandler {
  ev.preventDefault();
}

I could also use the function as a variable, like so, but it would force me to move the function before its usage:

const handleClicks: eventHandler = (ev: Event) => {
  ev.preventDefault();
}

browserEvents.onClicks(handleClicks);

so, is there a syntax to force a regular function to comply to a signature?

Answer 1

think you might be able to do something like this:

(but not sure what you're trying to return so I have typed it as void )

interface FunctionInterface {
  (e: eventHandler): void;
}

function f1(e): FunctionInterface {
  e.preventDefault();
}

const f2: FunctionInterface = e => {
  e.preventDefault()
};