如何在typescript中定义带有"this“上下文的函数

type MyFunctionType = (this: any, param1: string, param2: number) => void;

const myFunction: MyFunctionType = function(param1: string, param2: number) {
  // 在函数内部，可以使用this来引用该函数的调用对象
  console.log(this);
  console.log(param1, param2);
};

// 使用call()或apply()方法来指定调用对象
myFunction.call({ name: "John" }, "hello", 42);
myFunction.apply({ name: "Jane" }, ["world", 24]);

interface MyFunctionInterface {
  (this: any, param1: string, param2: number): void;
}

const myFunction: MyFunctionInterface = function(param1: string, param2: number) {
  console.log(this);
  console.log(param1, param2);
};

myFunction.call({ name: "John" }, "hello", 42);
myFunction.apply({ name: "Jane" }, ["world", 24]);