为什么这些 class 实现没有相同的行为？

Question

我试图为扩展 Function 的 class 编写概念证明，以演示可以基于另一个函数初始化的函数构造函数，并反映 this 和 arguments 以下列方式：

class Handle extends Function {
  constructor(functor) {
    super("functor", "slice", `
      "use strict";
      return functor
        .call(this, ...slice.call(arguments, 2));
    `);

    return this.bind(this, functor, Array.prototype.slice)
  }
}

let handle = new Handle(function test() {
  console.log(this instanceof Handle, arguments.length)
})

console.log(handle instanceof Handle, handle.length)

handle(1, 2, 3)

但是，根据我对 call 和 apply 的理解，我认为这会产生相同的行为：

class Handle extends Function {
  constructor(functor) {
    super("instance", "call", "slice", `
      "use strict";
      return call(this, instance, ...slice.call(arguments, 3));
    `);

    return Function.bind
      .call(this, functor, this, Function.call, Array.prototype.slice)
  }
}

let handle = new Handle(function test() {
  console.log(this instanceof Handle, arguments.length)
})

console.log(handle instanceof Handle, handle.length)

handle(1, 2, 3)

这抛出

Uncaught TypeError: call is not a function
  at Function.eval (eval at Handle (js:4), <anonymous>:5:14)

所以 call() 函数有问题。我的理解是，如果 call() 不是调用表达式的一部分，它的第一个参数将成为被调用的函数，其余参数将成为函数的上下文和参数，有点像 Function.call.call(this, instance, ...slice.call(arguments, 3)) 会做：

class Handle extends Function {
  constructor(functor) {
    super("instance", "call", "slice", `
      "use strict";
      return Function.call.call(this, instance, ...slice.call(arguments, 3));
    `);

    return Function.bind
      .call(this, functor, this, Function.call, Array.prototype.slice)
  }
}

let handle = new Handle(function test() {
  console.log(this instanceof Handle, arguments.length)
})

console.log(handle instanceof Handle, handle.length)

handle(1, 2, 3)

谁能解释一下我的误解，或者为什么情况似乎并非如此？

Answer 1

您正在调用 Function.call（我认为使用 Function.prototype.call 会更正确）而没有 this。不过，它需要一个，因为 this 值是它调用的函数。该问题的一个较短示例：

var call = Function.prototype.call;
call();

也许你打算 call.call(this, …)？

class Handle extends Function {
  constructor(functor) {
    super("instance", "call", "slice", `
      "use strict";
      return call.call(this, instance, ...slice.call(arguments, 3));
    `);

    return Function.bind
      .call(this, functor, this, Function.call, Array.prototype.slice)
  }
}

let handle = new Handle(function test() {
  console.log(this instanceof Handle, arguments.length)
})

console.log(handle instanceof Handle, handle.length)

handle(1, 2, 3)

Answer 2

Uncaught TypeError: call is not a function
so it's something wrong with the call() function.

请注意，不是 call 不是函数，而是 call 试图调用的东西。

My understanding was that if call() wasn't part of a call expression, its first argument would become the function being called

没有。被调用的函数始终是 call 调用的 this 上下文，当 call 未作为方法调用时，没有任何改变。你需要做 call.call(functor, context, ...args).

为什么这些 class 实现没有相同的行为？

Why don't these class implementations have identical behavior?

javascript

metaprogramming

function

call