新手求助

井井井1314

路过的师傅给看看 红色的两条是什么意义，有没有好像一样的

Public Function AddPlineSeg(ByVal ptSt As Variant, ByVal ptEn As Variant, ByVal width As Double) As AcadPolyline
    Dim objPline As AcadPolyline
    Dim ptArr(0 To 5) As Double

    ptArr(0) = ptSt(0)
    ptArr(1) = ptSt(1)
    ptArr(2) = ptSt(2)
    ptArr(3) = ptEn(0)
    ptArr(4) = ptEn(1)
    ptArr(5) = ptEn(2)

    Set objPline = ThisDrawing.ModelSpace.AddPolyline(ptArr)
    objPline.ConstantWidth = width
    objPline.Update
   
    Set AddPlineSeg = objPline
End Function

井井井1314

路过的老师 这个地方为什么是 (i\2)

Public Function AddPolygon(ByVal ptcen As Variant, ByVal number As Integer, ByVal radius As Date, Optional width As Double, Optional angle As Double = 0) As AcadLWPolyline
'定义动态数组
    Dim objPline As AcadLWPolyline
    Dim ptArr() As Double
    '顶点的个数为number,需要2*number个元素来表示
    ReDim ptArr(2 * number - 1)
   
    '每条边对应的角度
    Dim ang As Double
    ang = 2 * 3.1415926 / number
   
    '为点的坐标数组赋值
    Dim i As Integer
    For i = 0 To 2 * number - 1
        If i Mod 2 = 0 Then
            ptArr(i) = ptcen(0) + radius * Cos((i \ 2) * ang)
        ElseIf i Mod 2 <> 0 Then
            ptArr(i) = ptcen(1) + radius * Sin((i \ 2) * ang)
        End If
    Next i
        
    '创建多段线，并调整其特性
    Set objPline = AddLWPline(ptArr, width)
    objPline.Closed = True
    objPline.Rotate ptcen, angle
    objPline.Update
End Function

mikewolf2k

不加最后这句，function返回不了值。

mikewolf2k

井井井1314 发表于 2019-7-4 10:27
路过的老师 这个地方为什么是 (i\2)

Public Function AddPolygon(ByVal ptcen As Variant, ByVal number ...

数组的顶点列表是xy混在一起的，1/2是第一点的x/y, 3/4是第二点的x/y，所以要除以2取整才能获得相应的x/y。