TypeScript：如何从函数定义推导接口？

Absolutely, you can pull this off in TypeScript! The key thing to note is that when referencing a function in an interface, you need to use its type (not the function value itself). Let me walk you through a couple of clean ways to make this work correctly:

Method 1: Define a function type alias first #

This approach explicitly separates the function's signature from its implementation, making the type easy to reuse elsewhere:

// Step 1: Create a type alias for your function's signature
type MyFunctionType = (arg1: string, arg2: number) => boolean;

// Step 2: Implement the function using the type alias
const myFunction: MyFunctionType = (arg1, arg2) => {
  // Replace this with your custom logic
  return arg2 > arg1.length;
};

// Step 3: Build your interface using the type alias
interface MyInterface {
  someOtherProperty: string;
  myFunction: MyFunctionType; // Matches the exact signature you need
}

Method 2: Use typeof to reference an existing function's type #

If you already have the function defined, you can skip the type alias and use typeof to grab its signature directly:

// Step 1: Define your function first
function myFunction(arg1: string, arg2: number): boolean {
  // Your custom logic here
  return arg2 > arg1.length;
}

// Step 2: Reference the function's type in the interface with typeof
interface MyInterface {
  someOtherProperty: string;
  myFunction: typeof myFunction; // Inherits the exact signature from myFunction
}

Why your original approach doesn't work #

When you wrote myFunction: myFunction in the interface, TypeScript throws an error because the right-hand myFunction is a value (the actual function), not a type. Using typeof myFunction tells TypeScript to use the function's signature type instead of the value itself.

Example of using the interface #

Once everything is set up, you can create objects that conform to MyInterface like this:

const myObject: MyInterface = {
  someOtherProperty: "TypeScript is great!",
  myFunction: myFunction
};

// Test the function
console.log(myObject.myFunction("hello", 6)); // Logs true (since "hello" has length 5, 6 > 5)

内容的提问来源于stack exchange，提问作者Michael.Lumley