Question

    For j = 0 To 20
        For i = 0 To 20

            ReDim atoms(0 To 20, 0 To 20)
            ReDim atomschange(0 To 20, 0 To 20)
            atomschange(j, i) = 0

            If i Mod 2 = 0 And j Mod 2 = 0 Then
                [B2].Offset(j, i).Interior.ColorIndex = 37
                atoms(j, i) = 1
            Else
                [B2].Offset(j, i).Interior.ColorIndex = 36
                atoms(j, i) = 0
            End If
        Next i
    Next j
End Sub

我在做什么错了？

Answer 1

您将在重新初始化的循环内重设数组，然后将所有数据放入数组中。使用PRESERVE关键字来确保在重新命名数组时ReDim Preserve atoms(0 To 20, 0 To 20)
将数组重新排列为完全相同的大小没有任何意义。它们已经是0 To 20, 0 To 20，因为您已经设置了。将它们一次重新设置为for循环，然后不理会它们。
确保在代码顶部正确声明atoms和atomschange数组。它在抱怨（可能是在重新定义它们时）这两个变量不是数组。

在执行此代码之前，您已经知道了数组的大小，因此请声明它们并在声明中对其进行适当的调暗，然后再使用它们（删除循环中redim的多余和不正确的用法：

Sub somesumroutine()
    Dim atoms(0 to 20, 0 to 20) as Integer
    Dim atomschange(0 to 20, 0 to 20) as Integer

    For j = 0 To 20
        For i = 0 To 20             
            atomschange(j, i) = 0

            If i Mod 2 = 0 And j Mod 2 = 0 Then
                [B2].Offset(j, i).Interior.ColorIndex = 37
                atoms(j, i) = 1
            Else
                [B2].Offset(j, i).Interior.ColorIndex = 36
                atoms(j, i) = 0
            End If
        Next i
    Next j
End Sub

考虑使用UBound数组中的LBound和atoms来设置循环，这样就不必一遍又一遍地使用0 to 20。这样，您可以将数组变暗一次，而剩下的代码则不用担心：

Sub somesumroutine()
    Dim atoms(0 to 20, 0 to 20) as Integer
    Dim atomschange(0 to 20, 0 to 20) as Integer

    For j = lBound(atoms, 1) To uBound(atoms, 1)
        For i = lBound(atoms, 2) To uBound(atoms, 2)            
            atomschange(j, i) = 0

            If i Mod 2 = 0 And j Mod 2 = 0 Then
                [B2].Offset(j, i).Interior.ColorIndex = 37
                atoms(j, i) = 1
            Else
                [B2].Offset(j, i).Interior.ColorIndex = 36
                atoms(j, i) = 0
            End If
        Next i
    Next j
End Sub

预期数组，编译错误

1 个答案: