如何確保函數/方法參數是certian類型？

Question

是否可以確保功能/方法參數為某種類型？

例如，我有一個簡單的Character類，可以接受可選的Health對象。 是否可以檢查該參數是否為Health類型？ 當應用程序需要Health對象時，我不希望使用者傳遞一個整數。

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);
};

有任何想法嗎？

Answer 1

在這種特殊情況下，是的，您有兩個選擇：

1. instanceof ：

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

   從技術上講， instanceof檢查的是Health.prototype引用的對象是否在health的原型鏈中。

2. 檢查constructor

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

   注意，這很容易偽造： let a = {}; a.constructor = Health; let a = {}; a.constructor = Health;

通常，您可能希望使用前者，因為A）它允許Health子類型，並且B）當使用ES5和更早的語法進行繼承層次結構時， 許多人忘記了修復constructor ，並且最終指向了功能錯誤。

ES5語法示例：

 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

簡短的做法...

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