如何在高阶函数内部绑定函数引用?组合重于继承

Question

我正在尝试创建绑定每个函数对特定动物的引用，但它从未引用我期望的正确动物？

我

尝试使用各种 .call、.binds，但我不确定我是如何滥用它们的。

const sleeper = animal => ({
  sleep() {
    animal.energy += 10;
    console.log(`${animal.name} is sleeping! Energy is now ${animal.energy}!`);
  }
});
const eater = animal => ({
  eat() {
    animal.energy += 5;
    console.log(`${animal.name} is eating! Energy is now ${animal.energy}!`);
  }
});
const speaker = animal => ({
  speak() {
    animal.energy -= 3;
    console.log(`${animal.name} has uttered something!`)
  }
})

const Animal = (...args) => {
  return (name, energy = 0) => {
    const animal = {
      name,
      energy,
    }
    // this is where I am confused
    const boundFunctions = args.map(func => {
      return func.bind(animal)()
    })

    return Object.assign(animal, ...boundFunctions)
  }
}
const Monkey = Animal(sleeper, eater)
const Tiger = Animal(sleeper, eater, speaker)
const Reggie = Monkey("Reggie");
const Tony = Tiger('Tony')
Reggie.sleep()
Tony.eat()

我希望每个实例化的动物都能参考自己的名字和能量。但它是不确定的。

Answer 1

这里没有理由使用bind。行为函数只需要以动物作为参数调用，它们返回一个带有闭包的对象。用

const boundFunctions = args.map(func => func(animal))