2

我在玩功能装饰器。我想有一种方法可以轻松地装饰函数,而不必在声明函数的地方进行

var __slice = [].slice;

  this.around = function(decoration) {
    return function(base) {
      return function() {
        var argv, callback, __value__,
          _this = this;
        argv = 1 <= arguments.length ? __slice.call(arguments, 0) : [];
        __value__ = void 0;
        callback = function() {
          return __value__ = base.apply(_this, argv);
        };
        decoration.apply(this, [callback].concat(argv));
        return __value__;
      };
    };
  };

f("from outer");
function f(a){
  console.log("f OUTER",a);
};
f("from outer");
(function g(){
  f("from Inner (hoisted)");


function addDecorator(decoration, fn) {
  var decorated = around(decoration)(fn);
  eval(fn.name + " = decorated");
  decorated.unDecorate = function (){
    eval(fn.name + " = fn");
  }
}

  function myDecorator(cb){
      console.log("before");
      cb();
      console.log("after");
    }


  function f(a){
    console.log("f INNER",a);
  }

  f("from Inner (after declaration)");
  addDecorator(myDecorator, f);

  f("from Inner (Decorated)");
  f.unDecorate();
  f("from Inner (Undecorated)");
})();
f("from outer");

我希望能够addDecorator在里面的任何地方打电话g。即在f其声明上方悬挂的顶部。

在 Chrome 的控制台中,我得到以下输出:

f OUTER from outer
f OUTER from outer
f INNER from Inner (hoisted)
f INNER from Inner (after declaration)
before
f INNER from Inner (Decorated)
after
f INNER from Inner (Undecorated)
f OUTER from outer 
  1. 这可以在没有评估的情况下完成吗?
  2. addDecorator 可以移到 g 外吗?(没那么重要)
4

1 回答 1

2

如果你愿意稍微改变你的约定, 你可以消除eval和移动这个函数。g

这将具有相同的效果,但调用方式略有不同:

function addDecorator(decoration, fn) {
    var decorated = around(decoration)(fn);
    decorated.unDecorate = function () {
        return fn;
    }
    return decorated;
}

f = addDecorator(myDecorator, f);

f = f.unDecorate();

http://jsfiddle.net/8SuT9/1/

如果您不愿意更改调用/使用它的方式,那么恐怕您的两个问题的答案都是否定的。

于 2014-06-27T17:15:10.980 回答