Question

在以下代码中：

function outer() {
  var x = 'foo';

  function inner() {
    var y = x;     // y == 'foo'
    var x = 'bar'; // x == 'bar', y == undefined
  }
}

为什么变量y在inner()内未定义？它不应该引用x中的outer()吗？

如果删除了行var x = 'bar';，则y确实具有值'foo'。

Answer 1

int a = 5; auto *somePtr = 5; std::cout << typeid(*somePtr).name() << std::endl;函数被解释为它是这样编写的：

Sub TEST_SUB(surface)
    Application.ScreenUpdating = False
    Application.EnableEvents = False
    Application.Calculation = xlCalculationManual
    Worksheets("Sheet3").Activate
    ActiveSheet.DisplayPageBreaks = False

    Sheets("Sheet3").Range("A4:Z400").ClearContents


y = 4   'y is the row on sheet3 where we want to paste
For x = 4 To 20 'x is the current row from which we want to copy
        ' Decide if to copy based on whether the value in col 10 matches the parameter Surface
        ThisValue = Sheets("Tests_Master").Cells(x, 10).Value
        If ThisValue = surface Or x = 4 Then
            R1 = "A" + CStr(x) + ":K" + CStr(x) 'Range to copy from: row X columns 1-10

            'This next statement taks about 2 seconds to execute ! WHY????
            Sheets("Tests_Master").Range(R1).Copy Destination:=Sheets("sheet3").Range("A" + CStr(y))
            y = y + 1

        End If

Next x
Application.Calculation = xlCalculationAutomatic
Application.ScreenUpdating = True
Application.EnableEvents = True

End Sub

声明已悬挂，但初始化从上到下进行处理。因此，在整个inner函数中，符号function inner() { var y; var x; y = x; // y == undefined x = 'bar'; // x == 'bar' }和inner都引用该函数中声明为 local 的变量;特别是x是本地y，而不是封闭上下文中的x。因此，当评估x的初始化表达式时，y是本地x，尚未初始化; <{1>}的初始化程序后，的初始值设定项表达式为。

嵌套函数丢失变量引用

1 个答案: