你能在 TypeScript 中扩展一个函数吗？
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Can you extend a function in TypeScript?
提问by Will Munn
I help maintain a JavaScript library that produces spy functions which allow you to inspect how a passed-in function was called (mainly for use in unit testing).
我帮助维护一个 JavaScript 库，该库生成间谍函数，允许您检查传入函数的调用方式（主要用于单元测试）。
The library creates a function that has additional properties on it that allow you to inspect the calls.
该库创建了一个函数，该函数具有允许您检查调用的附加属性。
Is it possible to create a TypeScript definition that will allow the function to be passed in to methods that require a function AND have extra properties?
是否可以创建一个 TypeScript 定义，允许将函数传递给需要函数并具有额外属性的方法？
This is invalid, but something like:
这是无效的，但类似于：
class Spy extends function {
wasCalled: () => boolean;
...
}
Which would allow me to pass a spy into a function with this signature:
这将允许我将间谍传递到具有此签名的函数中：
function subjectUnderTest(callback:() => void) {
...
}
回答by Mattias Buelens
Yes, the TypeScript handbook calls this a "hybrid type", because it's a combination of a function type and a regular interface.
是的，TypeScript 手册将其称为“混合类型”，因为它是函数类型和常规接口的组合。
interface Spy {
(foo: string, bar: number) : boolean; // Just an example
wasCalled() : boolean;
}
var spy : Spy = createASpySomehow();
var result = spy("foo", 123);
if (spy.wasCalled()) {
// ...
}
回答by nalply
I wanted to extend a class with a function, too, and worked out a TypeScript-only solution. I am not really sure whether this is a good idea because clever solutions are not always good solutions. YMMV.
我也想用一个函数扩展一个类，并制定了一个仅限 TypeScript 的解决方案。我不确定这是否是一个好主意，因为聪明的解决方案并不总是好的解决方案。天啊。
Thanks to Mattias Buelens to provide a partial answer! I am building on it.
感谢 Mattias Buelens 提供部分答案！我正在建立它。
// same as in the answer of Mattias
interface Spy {
(foo: string, bar: number): boolean // Just an example
wasCalled(): boolean
}
// and now for the real solution!
class Spy {
_wasCalled: boolean
_baz: boolean // Just an example
private constructor(baz: boolean) {
this._wasCalled = false
this._baz = baz
}
wasCalled(): boolean {
return this._wasCalled
}
toString() { return '[object Spy]' }
static create(baz: boolean) {
const f = <Spy>function(this: Spy, foo: string, bar: number): boolean {
// Do your thing here. Use f instead of this!
console.log('wasCalled', f.wasCalled())
f._wasCalled = true
}
const spy = new Spy(baz)
Object.assign(f, spy)
Object.setPrototypeOf(f, Spy.prototype)
return f
}
}
The idea is to create a function and the instance of Spy, then assign both the prototype and the properties to the function. Return the instance from a static method. A bonus is the toString()method.
这个想法是创建一个函数和 的实例Spy，然后将原型和属性分配给函数。从静态方法返回实例。奖金是toString()方法。
const spy = Spy.create(true)
console.log('calling spy', spy('foo', 42))
console.log('instanceof', spy instanceof Spy)
works as expected.
按预期工作。
I don't think that new Spy()would work because we need to assign to a function not the other way round. And because we can't replace thiswe can't make thisa callable. A hypothetical way I see is to extend a class with really a function constructor somehow like this: class Spy2 extends function() {} {}, but I didn't find a way to get this working.
我认为这new Spy()行不通，因为我们需要分配给一个函数，而不是相反。而且因为我们不能替换，所以this我们不能做this一个可调用的。我看到的一种假设方法是使用真正的函数构造函数扩展一个类，就像这样: class Spy2 extends function() {} {}，但我没有找到让它工作的方法。
