Javascript 在Typescript中将类方法作为参数传递

Passing class method as parameter in Typescript

提问by xDreamCoding

I'm searching for a possibility to pass a class-method to a function which then can execute that function on an instance of that class. Something like that pseudocode: (note that this is an abstract example)

我正在寻找将类方法传递给函数的可能性，然后该函数可以在该类的实例上执行该函数。类似于那个伪代码：（注意这是一个抽象的例子）

class Foo {
    public somefunc() {
        // do some
    }
    public anyfunc() {
        // do any
    }
}

function bar(obj: Foo ,func: "Foo.method") {  // "that's what im looking for"
    obj.func();
}

bar(new Foo(), Foo.somefunc);  // do some
bar(new Foo(), Foo.anyfunc);  // do any

Is there a possiblity to do this?

有没有可能做到这一点？

I know i could be doing something like that:

我知道我可以做这样的事情：

class Foo {
    static somefunc(fooObj: Foo) {
        // do some
    }
    static anyfunc(fooObj: Foo) {
        // do any
    }
}

interface func {
    (fooObj: Foo);
}

function bar(obj: Foo, fn: func) {
    fn(obj);
}

bar(new Foo(), Foo.somefunc);  // do some
bar(new Foo(), Foo.anyfunc);  // do any

but that involves static functions which I don't want.

但这涉及我不想要的静态函数。

回答by Ryan Cavanaugh

This doesn't compile-time check that the function came from a Foo, but does the rest:

这不会在编译时检查函数是否来自 a Foo，但会检查其余部分：

class Foo {
    public somefunc() {
        // do some
    }
    public anyfunc() {
        // do any
    }
}

function bar(obj: Foo ,func: () => void) {
    func.call(obj);
}

bar(new Foo(), Foo.prototype.somefunc);  // do some
bar(new Foo(), Foo.prototype.anyfunc);  // do any

回答by Deilan

Typescript 2+ solution

打字稿 2+ 解决方案

TL;DR: TypeScript Playground, Repo with a demo

TL;DR: TypeScript Playground, Repo with a demo

Advantages:

好处：

1. Compile-time checking.
2. Won't let you lost thiscontext when passing an instance's method.
3. Don't lose performance: don't have to declare class' methods as instance methods (e. g. public somefunc = () => { return this.prop; }) - Learn more.
4. Don't mess with a class's prototype.
5. Consistent signature pattern: passing a callback as the first arg and thisArgas the second (e. g. Array.prototype.map()).

1. 编译时检查。
2. this传递实例的方法时不会让您丢失上下文。
3. 不要失去性能：不必将类的方法声明为实例方法（例如public somefunc = () => { return this.prop; }） -了解更多。
4. 不要弄乱类的原型。
5. 一致的签名模式：将回调作为第一个参数和thisArg第二个参数（例如Array.prototype.map()）传递。

Consider the following code:

考虑以下代码：

class Foo {
    private result: number = 42;

    public func(this: Foo): number {
        return this.result;
    }
}

function action(): void {
    console.log("Hello world!");
}

function bar(callbackFn: (this: void) => any, thisArg?: undefined): any;
function bar<T>(callbackFn: (this: T) => any, thisArg: T): any;
function bar<T, TResult>(callbackFn: (this: T) => TResult, thisArg: T): TResult {
    return callbackFn.call(thisArg);
}

const foo = new Foo();

bar(action); // success
bar(foo.func); // ERROR: forgot to pass `thisArg`
bar(foo.func, foo); // success

Turn your attention to the signature of Foo#func:

将注意力转移到以下签名上Foo#func：

public func(this: Foo): number

It states that this function should be invoked in a context of the class' instance. This is the first part of the solution which won't let you lost thiscontext.

它指出这个函数应该在类的实例的上下文中调用。这是解决方案的第一部分，不会让您失去this上下文。

The second part is barfunction overloads:

第二部分是bar函数重载：

function bar(callbackFn: (this: void) => any, thisArg?: undefined): any;
function bar<T>(callbackFn: (this: T) => any, thisArg: T): any;
function bar<T, TResult>(callbackFn: (this: T) => TResult, thisArg: T): TResult

This would let you you pass generic functions as well as instance methods.

这将使您可以传递通用函数以及实例方法。

You can learn more about these topics in TypeScript Handbook:

您可以在 TypeScript 手册中了解有关这些主题的更多信息：

1. thisparameters in callbacks
2. Function overloads
3. Generics

1. this回调中的参数
2. 函数重载
3. 泛型

回答by devi

Yes, declare function like this:

是的，像这样声明函数：

myfunction(action: () => void){
   action();
}

Call it like this from typescript:

从打字稿中这样称呼它：

myfunction(() => alert("hello"));

Or from javascript:

或者从javascript：

myfunction(function() { alert("hello"); });

Also you can pass method:

你也可以通过方法：

myfunction(this.someMethod);

回答by thetoast

I'm assuming you're looking for some way for the TypeScript compiler to enforce that the given function exists on Foo? Unfortunately, I don't think there's a way to do that. Maybe another TypeScript guru can come in here and answer that more concretely, but I'm pretty sure this is the closest that you can get:

我假设您正在寻找某种方式让 TypeScript 编译器强制执行给定的函数存在于 Foo 上？不幸的是，我认为没有办法做到这一点。也许另一位 TypeScript 大师可以进来并更具体地回答这个问题，但我很确定这是你能得到的最接近的：

class Foo {
    constructor(private name:string) { }

    public somefunc() {
        console.log("someFunc called on", this.name);
    }
    public anyfunc() {
        console.log("anyFunc called on", this.name);
    }
}

function bar(obj: Foo, func: string) {
    if (obj[func] && obj[func] instanceof Function) {
        obj[func]();
    } else {
        throw new Error("Function '" + func + "' is not a valid function");
    }
}

bar(new Foo("foo1"), "somefunc");  // output: 'somefunc called on foo1'
bar(new Foo("foo2"), "anyfunc");  // output: 'anyfunc called on foo1'
bar(new Foo("foo3"), "badFunction");  // throws: Error: Function 'badFunction' is not a valid function

回答by Clément MARTZLOFF

For my part; according to the statement of the problem, I could have done like this:

对我来说; 根据问题的陈述，我可以这样做：

class Foo {
    public constructor() {
        this.welcome = this.welcome.bind(this)
    }

    public welcome(msg: string): void {
        console.log(`hello ${msg}`)
    }
}

function bar(msg: string, fn: void): void {
    fn(msg)
}

const foo = new Foo()
bar('world', foo.welcome) // 'hello world'

In addition, I should point out that I was inspired by this clear explanation.

此外，我应该指出，我受到了这个清晰解释的启发。

Hope it helps !

希望能帮助到你 ！

回答by Erik Lieben

Javascript would allow this, but not sure if that is what you want?

Javascript 会允许这样做，但不确定这是否是您想要的？

class Foo {
 public someFunc(name:string){
  return "Hello, " + name;
 }

function bar(funcName: string) {
    return eval(funcName);
}

console.log(bar("new Foo().someFunc('erik')"));