Javascript更改绑定值

Question

此功能用于报告和增加每个特定障碍物的总数（计数器）。

function warningMaker( obstacle ){ 
      var count =  0;
      return function ( number, location ) {
        count++;
        alert("Beware! There have been " +
              obstacle +
              " sightings in the Cove today!\n" +
              number +
              " " +
              obstacle +
              "(s) spotted at the " +
              location +
              "!\nThis is Alert #" + count + " today for " + obstacle + " danger."
             );
      };
    }

现在如果我通过说例如

来调用该函数

warningMaker("obstacleName1")(2,"locationName");   ===counter 1
warningMaker("obstacleName1")(2,"locationName");   ===counter still 1

但如果我这样称呼它，

var obstacle1Maker = warningMaker("obstacleName1");
var obstacle2Maker = warningMaker("obstacleName2");
obstacle1Maker(2,"MiddleEarth");
obstacle1Maker(2,"Hogwarts");
obstacle2Maker(3,"Narnia");

计数器针对每个特定障碍物递增，这怎么可能？我是Javascript的新手，我试图掌握这样的事情背后的概念。

Answer 1

两个不同的闭包装有两个不同的计数器。你想要这样：

var obstacles={};
function warningMaker( obstacle ){ 
  var obstacles[obstacle] =  obstacles[obstacle]||0;
  return function ( number, location ) {
    obstacles[obstacle]++;
    alert("Beware! There have been " +
          obstacle +
          " sightings in the Cove today!\n" +
          number +
          " " +
          obstacle +
          "(s) spotted at the " +
          location +
          "!\nThis is Alert #" + obstacles[obstacle] + " today for " + obstacle + " danger."
         );
  };
}

我将所有计数存储在全局对象中。因此它不依赖于封闭等......

为什么您的代码无法正常工作：

warningMaker("obstacleName1");

这组装了一个新的函数上下文（function + closure）。将函数移出其范围（变量周围）时会发生这种情况。该函数仍然可以访问它，因为它绑定到它们（称为闭包）。因此，如果你再调用两次，你将创建两个不同的上下文，其中包含两个不同的 count 变量......