Question

是否可以确保函数/方法参数具有某种类型？

例如，我有一个简单的Character类，它接受一个可选的Health对象。是否可以检查该参数是否为Health类型？当应用程序需要Health对象时，我不希望使用者传入整数。

let Character = function(health) {
    if(typeof health === 'undefined')
        this.health = new Health(100);
    else
        this.health = health;
};

Character.prototype.hit = function(hitPoints) {
    this.health.subtract(hitPoints);
};

有什么想法吗？

Answer 1

在这种特殊情况下，是的，您有两种选择：

instanceof：
```
if (health instanceof Health) {
    // It's a Health object *OR* a derivative of one
}
```
从技术上讲，instanceof检查的对象是Health.prototype引用的对象是health的原型链。

检查constructor

if (health.constructor === Health) {
    // Its `constructor` is `Health`, which usually (but not necessarily)
    // means it was constructed via Health
}

请注意，这很容易伪造：let a = {}; a.constructor = Health;

通常你可能想要达到前者，因为A）它允许Health的子类型，和B）当使用ES5和更早的语法进行继承层次结构时，很多人们忘记修复constructor并最终指向错误的功能。

ES5语法中的示例：

var Health = function() {
};

var PhysicalHealth = function() {
  Health.call(this);
};
PhysicalHealth.prototype = Object.create(Health.prototype);
PhysicalHealth.prototype.constructor = PhysicalHealth;

var h = new PhysicalHealth();
log(h instanceof Health);     // true
log(h.constructor == Health); // false

function log(msg) {
  var p = document.createElement('p');
  p.appendChild(document.createTextNode(msg));
  document.body.appendChild(p);
}

或使用ES2015（ES6）：

class Health {
}

class PhysicalHealth extends Health {
}

let h = new PhysicalHealth();
log(h instanceof Health);     // true
log(h.constructor == Health); // false

function log(msg) {
  let p = document.createElement('p');
  p.appendChild(document.createTextNode(msg));
  document.body.appendChild(p);
}

Answer 2

简短的做法...

let character = function(health) { this.health = (health instanceof Health) ? health : new Health(100); }

如何确保函数/方法参数是certian类型？

2 个答案: