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我一直在尝试解决这个问题，但无法解决。我有以下 VBA 代码。运行时出现“运行时错误 6：溢出”。我想我已经为每个变量分配了正确的类型。请帮忙！！

Sub UseCoeff()
Dim a, b As Long
Dim Value1 As Double

ThisWorkbook.Sheets("UseTableBEA").Activate
For b = 2 To 427
    For a = 2 To 431

        Value1 = ThisWorkbook.Sheets("UseTableBEA").Cells(a, b).Value / ThisWorkbook.Sheets("UseTableBEA").Cells(432, b).Value
        ThisWorkbook.Sheets("UseCoeff").Cells(a, b).Value = Value1

    Next a
Next b
End Sub

每个单元格的结果可能介于 0 和 1 之间，最多八位小数。提前致谢！！

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2 回答 2

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如果ThisWorkbook.Sheets("UseTableBEA").Cells(432, b).Value曾经为 0，那么如果也是 0，ThisWorkbook.Sheets("UseTableBEA").Cells(a, b).Value / ThisWorkbook.Sheets("UseTableBEA").Cells(432, b).Value则会导致溢出错误ThisWorkbook.Sheets("UseTableBEA").Cells(a, b).Value。

为避免这种情况，请使用 if 语句过滤掉 0，只需使用您想要的实际逻辑更新注释：

If ThisWorkbook.Sheets("UseTableBEA").Cells(432, b).Value = 0 Then
    Value1 = 'Whatever you want but probably 0.
Else
    Value1 = ThisWorkbook.Sheets("UseTableBEA").Cells(a, b).Value / ThisWorkbook.Sheets("UseTableBEA").Cells(432, b).Value
End If
于 2012-09-18T04:18:43.930 回答
2

Dim a, b As Long

a 未声明为 long 它被声明为变体试试这个：

Dim a as long, b as Long

编辑： 经过测试，它仍然会产生堆栈溢出错误。但是，正确声明变量是一种很好的做法。

您也不需要这一行：

ThisWorkbook.Sheets("Sheet1").Activate

如果你声明UseTableBEA你可以精简你的代码：

Dim Ws1 as Worksheet
Set Ws1 = Sheets("UseTableBEA")

然后Ws1，您可以在以前使用过的地方使用ThisWorkbook.Sheets("UseTableBEA")

于 2012-09-18T12:47:48.523 回答