如何将自定义类型的数组作为参数传递给过程?

mikewolf2k · 发表于 2004-3-8 21:15:00

Type aa
   name As String
End Type
Sub test()
   Dim a() As aa    'dim a()
   ReDim a(1)
   Call addtype(a, "ok")
End Sub
Sub addtype(a, str As String)
   Dim i As Integer
   ReDim Preserve a(UBound(a) + 1)
   With a(UBound(a))
   .name = str
   End With
End Sub
请看看应该如何实现标题所述功能?
这段代码,我试过了,如果将a()定义为aa就会出错,不定义为aa就可以通过(a.name语句相应调整)

mikewolf2k · 发表于 2004-3-10 20:46:00

救命啊,就要沉了!!!

莫名 · 发表于 2004-3-10 21:15:00

看下面主题我的最后复贴

mikewolf2k · 发表于 2004-3-10 21:27:00

莫大,看过了,现在问题是:"如果将a()定义为aa就会出错,不定义为aa就可以通过(a.name语句相应调整)"

my_computer · 发表于 2004-3-10 21:46:00

我改了，现在可以了
Public Type aa
   name As String
End Type
Sub test()
   Dim a() As aa    'dim a()
   ReDim a(1)
   addtype a, "ok"
End Sub
Sub addtype(a() As aa, str As String)
   Dim i As Integer
   ReDim Preserve a(UBound(a) + 1)
   With a(UBound(a))
   .name = str
   End With
End Sub

莫名 · 发表于 2004-3-10 22:03:00

[I]注意a要定义为全局变量或窗体的模块级变量（下面画线的语句）。 [/I]Private Type aa
   name As String
   firstrow As Integer
   lastrow As Integer
End Type
[U]Private a() As aa[/U]

Private Sub addtype(str)
   Dim k As Integer
   k = UBound(a, 1)
   ReDim Preserve a(k + 1)
   k = UBound(a, 1)
   With a(k)
   .name = str
   .firstrow = a(k - 1).lastrow
   .lastrow = .firstrow
   End With
   Debug.Print k, a(k).name, a(k).firstrow, a(k).lastrow

End Sub
Private Sub Command1_Click()

   ReDim a(0)
   a(0).firstrow = 2
   a(0).lastrow = 2
   a(0).name = "ok"
   Dim i As Integer

   addtype "ok"
End Sub

mikewolf2k · 发表于 2004-3-10 22:06:00

5楼的方法解决,谢谢

如何将自定义类型的数组作为参数传递给过程?

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