在Typescript中传递类方法作为参数 [英] Passing class method as parameter in Typescript

问题描述

我正在寻找将类方法传递给一个函数的可能性，然后该函数可以在该类的一个实例上执行该函数。
类似于伪代码：（注意这是一个抽象的例子）

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)

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

有没有可能做到这一点？

Is there a possiblity to do this?

我知道我可以这样做：

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

但涉及我不想要的静态函数。

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

推荐答案

这不会编译时检查函数是否来自 Foo ，但其余的：

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

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

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