使用扩展原型定义可调用的TypeScript函数(Define a callable TypeScript function with extended prototype)

 我想定义一个可调用函数foo() （不是一个类，我不想在这里使用new运算符），它也会有一个属性foo.bar() 。 在普通的JS中，它将如下所示： 
 
function foo() {
    // ...
}

foo.prototype.bar = function bar() {
    // ...
}
 
 我在TypeScript中使用这样的接口尝试了它： 
 
interface IFoo {
    (): any;
    bar(): any;
}

const foo: IFoo = function(): any {
    // ...
};

foo.prototype.bar = function bar(): any {
    // ...
};
 
 但是我得到一个错误： 
 
error TS2322: Type '() => any' is not assignable to type 'IFoo'.
 
 似乎TS抱怨定义foo和它的属性bar之间的中间状态，即由于foo还没有属性bar它不能被分配给IFoo类型的const 。 我怎样才能解决这个问题？ 
 
 以另一种方式提问：我如何为IFoo提供实施？ 

I want to define a callable function foo() (not a class! I don't want to use the new operator here) that will also have a property foo.bar(). In plain JS it will look like this:
 
function foo() {
    // ...
}

foo.prototype.bar = function bar() {
    // ...
}
 
I tried it in TypeScript using an interface like this:
 
interface IFoo {
    (): any;
    bar(): any;
}

const foo: IFoo = function(): any {
    // ...
};

foo.prototype.bar = function bar(): any {
    // ...
};
 
But I get an error:
 
error TS2322: Type '() => any' is not assignable to type 'IFoo'.
 
It seems that TS complains about the intermediate state between defining foo and its property bar, that since foo does not yet have a property bar it cannot be assigned to a const of type IFoo. How can I fix this?
 
To ask it in another way: how can I provide an implementation for IFoo?

原文：https://stackoverflow.com/questions/43613387