如何确保函数/方法参数是certian类型？

Question

是否可以确保功能/方法参数为某种类型？

例如，我有一个简单的Character类，可以接受可选的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，我不希望使用者传递一个整数。

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);
};

有任何想法吗？

Answer 1

在这种特殊情况下，是的，您有两个选择：

1. instanceof ：

    if (health instanceof Health) { // It's a Health object *OR* a derivative of one } 

   从技术上讲， instanceof检查的是Health.prototype引用的对象是否在health的原型链中。

2. 检查constructor

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

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

通常，您可能希望使用前者，因为A）它允许Health子类型，并且B）当使用ES5和更早的语法进行继承层次结构时， 许多人忘记了修复constructor ，并且最终指向了功能错误。

ES5语法示例：

 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); }