Question

刚开始学习闭包并在FireBug中编写了这段代码：

var later;

function outerFunc() {
  var innerVar = "Inside Outer";
  function innerFunc() {
     console.log(innerVar);
  }

  later = innerFunc;
};

outerFunc();
later();

但是在调用later();时，它返回一个未定义的值。不应该打印"Inside Outer" ?

Answer 1

var later;

function outerFunc() {
  var innerVar = "Inside Outer";
  function innerFunc() {
     console.log(innerVar);
  }

  later = innerFunc;
};

outerFunc(); //Don't run innerFunc, just set later = innerFunc
later(); // Execute innerFunc

所以它只记录了一次..看起来很不错..

关闭不在函数范围之外工作

1 个答案: