javascript - 在 JavaScript 中设置对象值

Question

我已经尽力解决了这个问题，现在我卡住了，为什么第四个警报返回未定义？

function buttonClick()
{
  var myTest = function()
  {
    var _setDirectlyInside = "So far so good...";
    var _setInFunctionCalledInside;
    var _setInFunctionCalledFromOutside;

    (function(){
      _setInFunctionCalledInside = "This would indicate scope isn't the problem?";
    })();

    return {
      basic : "Easy stuff",
      setDirectlyInside : _setDirectlyInside,
      setInFunctionCalledInside : _setInFunctionCalledInside,
      functionCallFromOutside : function(){
        _setInFunctionCalledFromOutside = "Why does this come back as undefined?";
      },
      setInFunctionCalledFromOutside : _setInFunctionCalledFromOutside
    }
  };

  var test = myTest();

  alert(test.basic); // Returns "Easy stuff"
  alert(test.setDirectlyInside); // Returns "So far so good..."
  alert(test.setInFunctionCalledInside); // Returns "This would indicate scope isn't the problem?"

  test.functionCallFromOutside();
  alert(test.setInFunctionCalledFromOutside); // Broken, returns undefined
}

解决方案:

setInFunctionCalledFromOutside : _setInFunctionCalledFromOutside, // Won't work
setInFunctionCalledFromOutsideGetter : function(){
    return _setInFunctionCalledFromOutside; // Will work
}

...

alert(test.setInFunctionCalledFromOutside); // Broken, returns undefined
alert(test.setInFunctionCalledFromOutsideGetter()); // Now works

Answer 1

这个:

return {
  basic : "Easy stuff",
  setDirectlyInside : _setDirectlyInside,
  setInFunctionCalledInside : _setInFunctionCalledInside,
  functionCallFromOutside : function(){
    _setInFunctionCalledFromOutside = "Why does this come back as undefined?";
  },
  setInFunctionCalledFromOutside : _setInFunctionCalledFromOutside
}

...不会导致 setInFunctionCalledFromOutside 始终返回与 _setInFunctionCalledFromOutside 相同的值。相反，_setInFunctionCalledFromOutside 将在 return 语句执行时计算，其值将放在 setInFunctionCalledFromOutside 中。因此，functionCallFromOutside() 不会影响 setInFunctionCalledFromOutside。