Excel中的VBA函数可以返回范围吗？

Question

在尝试执行以下操作时,我似乎遇到类型不匹配错误:

在新工作簿中:

A1 B1
5  4

Function Test1() As Integer
    Dim rg As Range
    Set rg = Test2()
    Test1 = rg.Cells(1, 1).Value
End Function
Function Test2() As Range
    Dim rg As Range
    Set rg = Range("A1:B1")
    Test2 = rg
End Function

Adding = Test1()应返回5,但代码似乎在从test2()返回范围时终止.是否可以返回范围？

Answer 1

范围是一个对象.分配对象需要使用SET关键字,看起来你忘记了Test2函数中的一个:

Function Test1() As Integer
    Dim rg As Range
    Set rg = Test2()
    Test1 = rg.Cells(1, 1).Value
End Function

Function Test2() As Range
    Dim rg As Range
    Set rg = Range("A1:B1")
    Set Test2 = rg         '<-- Don't forget the SET here'
End Function

Answer 2

您还可以返回Variant()表示值数组的a.下面是一个函数示例,它将值从一个范围反转到一个新范围:

Public Function ReverseValues(ByRef r_values As Range) As Variant()
    Dim i As Integer, j As Integer, N As Integer, M As Integer
    Dim y() As Variant
    N = r_values.Rows.Count
    M = r_values.Columns.Count
    y = r_values.value    'copy values from sheet into an array
    'y now is a Variant(1 to N, 1 to M) 
    Dim t as Variant
    For i = 1 To N / 2
        For j = 1 To M
            t = y(i, j)
            y(i, j) = y(N - i + 1, j)
            y(N - i + 1, j) = t
        Next j
    Next i

    ReverseValues = y
End Function

在工作表中,您必须将此函数应用为数组公式(带Ctrl- Shift- Enter),并选择适当数量的单元格.Swap()函数的细节在这里并不重要.

请注意,对于许多行,这非常有效.执行x = Range.Value和Range.Value = x操作何时x是一个数组,并且范围包含多行列比直接在单元格上逐个执行操作要快许多倍.

Answer 3

将 Test2 中的最后一行更改为：

Set Test2 = rg