覆盖函数

Question

以下“aMacro”返回错误“检测到模糊名称”我理解为什么。任何人都知道一种覆盖第一个定义的方法，并且只使用函数内部的定义，以便aFunction将返回x - y？

除了更改名称外。

Function aFunction(x As Integer, y As Integer) As Integer

    aFunction = x + y

End Function



Sub aMacro()

    Function aFunction(x As Integer, y As Integer) As Integer
        aFunction = x - y
    End Function
    MsgBox aFunction(4, 3)

End Function

Answer 1

一个函数可以是Private或Public，但范围始终是整个模块。

Answer 2

这可以模拟＆＃34;覆盖功能＆＃34;有4个课程模块： 函数，IFunction，FunctionAdd，FunctionSubtract。

类模块功能：

Function aFunction(x As Integer, y As Integer) As Integer
    aFunction = x + y
End Function

接口IFunctions：

Function aFunction(x As Integer, y As Integer) As Integer
End Function

类模块FunctionAdd：

Implements IFunctions

Private mFunctions As Functions

Private Sub Class_Initialize()
    Set mFunctions = New Functions
End Sub

Private Sub Class_Terminate()
    Set mFunctions = Nothing
End Sub

Private Function IFunctions_aFunction(x As Integer, y As Integer) As Integer
    IFunctions_aFunction = mFunctions.aFunction(x, y) ' Uses the standard aFunction
End Function

类模块FunctionSubtract：

Implements IFunctions

Private mFunctions As Functions

Private Sub Class_Initialize()
    Set mFunctions = New Functions
End Sub

Private Sub Class_Terminate()
    Set mFunctions = Nothing
End Sub

Private Function IFunctions_aFunction(x As Integer, y As Integer) As Integer
    IFunctions_aFunction = x - y ' Override aFunction, subtract values
End Function

你可以用这个来测试：

Dim f As IFunctions
Set f = New FunctionAdd: Debug.Print f.aFunction(1, 2)
Set f = New FunctionSubtract: Debug.Print f.aFunction(1, 2)

当然，这对于一个功能来说是乏味的。 我可能很有用，你可以在很多类中覆盖很多函数。

Answer 3

尝试添加函数中的Optional值。如果调用中未包含可选值，则它不会是函数中的引用。

Function aFunction(x As Integer, y As Integer, Optional override As Boolean) As Integer
    If Not override Then
        aFunction = x + y
    Else
        aFunction = x - y
    End If
End Function
Sub aMacro()
    MsgBox aFunction(4, 3)
    MsgBox aFunction(4, 3, True)
End Sub