Ms access 函数不返回屏幕分辨率值

函数工作并获得屏幕分辨率。但当我试着回到sub时，fun_x和fun_y都是空的。为什么?

Private Sub Form_Open(Cancel As Integer)

Dim sub_x As Long, sub_y As Long

ScreenRes fun_x:=sub_x, fun_y:=sub_y

Debug.Print sub_x, sub_y, fun_x, fun_y

End Sub

---

模块名称：MOD_GET_RES

Option Compare Database

Declare Function GetSystemMetrics32 Lib "User32" _
Alias "GetSystemMetrics" (ByVal nIndex As Long) As Long

Function ScreenRes(ByVal fun_x As Long, ByVal fun_y As Long)

fun_x = GetSystemMetrics32(0) ' width in points
fun_y = GetSystemMetrics32(1) ' height in points

End Function

---

您已经传递了函数参数

ByVal

（ue），因此当它返回时，只修改一个副本。隐式或不隐式使用

ByRef

（参照）使其工作：

Function ScreenRes(fun_x As Long, fun_y As Long)

您还可以将

GetSystemMetrics32

（为什么要使用更长的别名？）设为私有

编辑：下面是一个小例子的更多解释

Option Explicit

Sub CantChange(ByVal a As Integer, ByVal b As Integer)
    'When this gets called, a copy of the original variables
    'is pushed on the stack.
    a = a * 2 'The copy is altered.
    b = b + 1
    'When leaving this Sub, the copy is discarded.
End Sub

'exactly the same Sub procedure... except for the (implicit) ByRef
Sub Change(a As Integer, b As Integer)
    'When this gets called, references
    '(i.e. the location of the original variable)
    'are pushed on the stack.
    a = a * 2 'The original variable is altered.
    b = b + 1
End Sub

Sub Test()
    Dim a As Integer, b As Integer
    a = 1
    b = 5
    Debug.Print "before:", a, b
    CantChange a, b
    Debug.Print "unchanged:", a, b
    Change a, b
    Debug.Print "changed:", a, b
End Sub

---

您已经传递了函数参数

ByVal

（ue），因此当它返回时，只修改一个副本。隐式或不隐式使用

ByRef

（参照）使其工作：

Function ScreenRes(fun_x As Long, fun_y As Long)

您还可以将

GetSystemMetrics32

（为什么要使用更长的别名？）设为私有

编辑：下面是一个小例子的更多解释

Option Explicit

Sub CantChange(ByVal a As Integer, ByVal b As Integer)
    'When this gets called, a copy of the original variables
    'is pushed on the stack.
    a = a * 2 'The copy is altered.
    b = b + 1
    'When leaving this Sub, the copy is discarded.
End Sub

'exactly the same Sub procedure... except for the (implicit) ByRef
Sub Change(a As Integer, b As Integer)
    'When this gets called, references
    '(i.e. the location of the original variable)
    'are pushed on the stack.
    a = a * 2 'The original variable is altered.
    b = b + 1
End Sub

Sub Test()
    Dim a As Integer, b As Integer
    a = 1
    b = 5
    Debug.Print "before:", a, b
    CantChange a, b
    Debug.Print "unchanged:", a, b
    Change a, b
    Debug.Print "changed:", a, b
End Sub

---

这起作用了，但我不明白为什么。我想我需要做一些关于差异的阅读。我添加了一个简单的例子，我希望现在更清楚；）谢谢你花时间把这些整理好。很有助于理解为什么，而不是如何。这是有效的，但我不明白为什么。我想我需要做一些关于差异的阅读。我添加了一个简单的例子，我希望现在更清楚；）谢谢你花时间把这些整理好。非常有助于理解为什么而不是如何。