VB关于矩阵的操作问题

@H_404_0@简单的代码,在百度回答的时候顺便写下,可能有人会重复的问到

@H_404_0@添加一个Form,Paste如下代码即可看效果

@H_404_0@

Option Explicit

Private Type MATRIX
    nRowCount As Long
    nColCount As Long
    pMatData() As Double
End Type

Private Sub Form_Load()
    '[11 12 13]
    '[21 22 23]
    '[31 32 33]
    Dim d As Double,i As Long,j As Long
    Dim m As MATRIX,s As String
    
    s = "11 12 13" & vbCrLf & "21 22 23" & vbCrLf & "31 32 33"
    InitMatrix m,s,3,Asc(vbCrLf),Asc(" ")
    d = GetMaxItem(m,i,j)
    Msg<a href="https://www.jb51.cc/tag/Box/" target="_blank" class="keywords">Box</a> "最大值: " & CStr(d) & " 位置是 第 " & CStr(i) & " 行,第 " & CStr(j) & " 列." & vbCrLf & "矩阵原始数据:" & vbCrLf & s & vbCrLf
    'End
End Sub

Private Sub InitMatrix(lpMatrix As MATRIX,lpString As String,ByVal nRow As Long,ByVal nCol As Long,Optional ByVal RowChar As Long = &HD,Optional ByVal ColChar = &H20)
    Dim strLine() As String,strCol() As String
    Dim cRow As String * 1,cCol As String * 1
    Dim i As Long,j As Long
    Dim n As Long
    
    If nRow * nCol = 0 Then Exit Sub
    lpMatrix.nRowCount = nRow
    lpMatrix.nColCount = nCol
    ReDim lpMatrix.pMatData(0 To nRow - 1,0 To nCol - 1)
    
    If lpString = "" Then Exit Sub
    If RowChar * ColChar = 0 Then Exit Sub
    cRow = Chr(RowChar)
    cCol = Chr(ColChar)
    strLine = Split(lpString,cRow)
    n = UBound(strLine)
    
    On Local Error Resume Next
    For i = 0 To n
        strCol = Split(strLine(i),cCol)
        For j = 0 To UBound(strCol)
            lpMatrix.pMatData(i,j) = strCol(j)
        Next j
    Next i
End Sub

Private Function GetMaxItem(lpMatrix As MATRIX,nRow As Long,nCol As Long) As Double
    Dim i As Long,j As Long
    Dim d As Double
    For i = 0 To lpMatrix.nRowCount - 1
        For j = 0 To lpMatrix.nColCount - 1
            If lpMatrix.pMatData(i,j) > d Then
                d = lpMatrix.pMatData(i,j)
                nRow = i + 1
                nCol = j + 1
            End If
        Next j
    Next i
    GetMaxItem = d
End Function

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VB关于矩阵的操作问题

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