Question

我有以下一段JavaScript代码

function freezeProps(o) {
var props = (arguments.length == 1) // If 1 arg
? Object.getOwnPropertyNames(o) // use all props
: Array.prototype.splice.call(arguments, 1); // else named props
props.forEach(function(n) { // Make each one read-only and permanent
// Ignore nonconfigurable properties
if (!Object.getOwnPropertyDescriptor(o,n).configurable) return;
Object.defineProperty(o, n, { writable: false, configurable: false });
});
   return o; // So we can keep using it
}

// Make the named (or all) properties of o nonenumerable, if configurable.
  function hideProps(o) {
  var props = (arguments.length == 1) // If 1 arg
  ? Object.getOwnPropertyNames(o) // use all props
  : Array.prototype.splice.call(arguments, 1); // else named props
   props.forEach(function(n) { // Hide each one from the for/in loop
  // Ignore nonconfigurable properties
   if (!Object.getOwnPropertyDescriptor(o,n).configurable) return;
   Object.defineProperty(o, n, { enumerable: false });
  });
     return o;
   }


 function Range(from, to) { // Constructor for an immutable Range class
      this.from = from;
      this.to = to;
       freezeProps(this); // Make the properties immutable
 }

Range.prototype = hideProps({ // Define prototype with nonenumerable   properties
 constructor: Range,
 includes: function(x) { return this.from <= x && x <= this.to; },
  foreach: function(f) {for(var x=Math.ceil(this.from);x<=this.to;x++)    f(x);},
  toString: function() { return "(" + this.from + "..." + this.to + ")"; }
 });

  var range = new Range(10, 12);
  var result = range.includes(10.9666);
  alert(result);

我在函数freezeProps（）和hideProps（）中无法理解的是，我可以将多个参数传递给hideProps（arg1，arg2，arg3）然后

  arguments.length != 1

将执行以下代码行

 var props =  Array.prototype.splice.call(arguments, 1);

但是道具将包含像[arg2，arg3]这样的结果数组，如果是这样的话，那么为什么仍然可以将对象arg1上的命名属性（从代码中的注释中清楚地看到）更改为不可数。这对我来说没有意义。任何建议都将受到高度赞赏。

通过实用程序功能更改属性属性

0 个答案: