我的想法是创建一个函数来传递双倍数组,如下所示:
Function pass(a() As Double, b() as double) As Boolean
Dim i As Integer, j As Integer
ReDim a(0 To UBound(b, 1), 0 To UBound(b, 2))
For i = 0 To UBound(a, 1)
For j = 0 To UBound(a, 2)
a(i, j) = b(i, j)
Next
Next
End Function
这样人们就可以使用:
pass(a,b)
然后是a=b。
a=b
不起作用..(哦,天哪,我好怀念matlab。)
好了,回到问题上:
这对于两个双精度数组很有效,但是如果我得到一个变量,当我使用这个变量时,我可以得到一个变量,这不起作用。那么有没有一种将变量值传递给双数组的解决方案呢?数据的结构类似于双精度数组,但不是double()类型。
发布于 2018-08-22 03:51:47
为了解决这个问题,我将删除参数的类型。
Sub test()
Dim VarArry(0 To 4, 0 To 4) As Variant
Dim DblArry(0 To 4, 0 To 4) As Double
Dim VarAssing() As Variant
Dim DblAssing() As Double
Dim c As Boolean
VarArry(0, 1) = 34
DblArry(0, 1) = 34
c = pass(VarAssing, VarArry)
Debug.Print (TypeName(VarAssing))
Debug.Print (VarAssing(0, 1))
c = pass(VarAssing, DblArry)
Debug.Print (TypeName(VarAssing))
Debug.Print (VarAssing(0, 1))
c = pass(DblAssing, VarArry)
Debug.Print (TypeName(DblAssing))
Debug.Print (VarAssing(0, 1))
c = pass(DblAssing, DblArry)
Debug.Print (TypeName(DblAssing))
Debug.Print (VarAssing(0, 1))
End Sub
Function pass(a, b) As Boolean
Dim i As Integer, j As Integer
ReDim a(0 To UBound(b, 1), 0 To UBound(b, 2))
For i = 0 To UBound(a, 1)
For j = 0 To UBound(a, 2)
a(i, j) = b(i, j)
Next
Next
End Function
输出如下所示
Variant()
34
Variant()
34
Double()
34
Double()
34
https://stackoverflow.com/questions/51948918
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