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'一个矩形的形状，用以下表示
dim str as string
str="30,-70|50,-70|50,-84|30,-84"
dim jl as long '输入一个值，在str矩形的内部画一个xy坐标都内移jl值的矩形，这个矩形正好套在上个乱形之内。
新得到的矩形的坐标值是"32,-72|48,-72|48,-82|32,-82"
我用的方法是将每个顶点可能发生的四种情况都调用ptinpoly函数判断如果返回0值表示在内部，但是四个值一个都不返回0值，下面的函数是不是有错误啊，高手能不能帮我检查一下，看看有什么问题，或是谁有VB或VBA的代码发表一下，急呀，检查不出问题呀。
它的四种情况是：
x-,y-
x+,y+
x-,y+
x+,y-
这四种情况中应该有一种是在内部的
'判断给定点 pt 是否在多边形 poly 内
'返回 0 在内部，-1 在外面
'返回 > 0 表示点在第几条有向线段上
Private Function PtInPoly(ByVal pt As Variant, ByVal poly As Variant) As Integer
Dim i As Integer
Dim status, lastStauts As Integer
Dim cnt As Integer
Dim POS, temp As Integer
Dim var_poly As Variant
Dim var_polyi As Variant
var_poly = Split(poly(1), ",")
cnt = 0
lastStauts = IIf(var_poly(1) > pt(1), 1, IIf(var_poly(1) = pt(1), 0, -1))
For i = 1 To UBound(poly)
var_poly = Split(poly(i), ",")
status = IIf(var_poly(1) > pt(1), 1, IIf(var_poly(1) = pt(0) And pt(0) >= var_poly(0)) And (var_polyi(1) = pt(1) And pt(1) >= var_poly(1))) Then
PtInPoly = i
Exit Function
End If
' 跨越
If (temp > 0 And POS = 1 Or temp
'判断点在线的哪侧
Private Function SideOfLine(ByVal p1 As Variant, ByVal p2 As Variant, ByVal pt As Variant) As Integer
Dim RR, TOP, LL As Integer
RR = -1
TOP = 0
LL = 1
Dim c1 As Double
Dim c2 As Double
Dim var_p1 As Variant
Dim var_p2 As Variant
var_p1 = Split(p1, ",")
var_p2 = Split(p2, ",")
c1 = (var_p2(0) - pt(0)) * (pt(1) - var_p1(1))
c2 = (var_p2(1) - pt(1)) * (pt(0) - var_p1(0))
SideOfLine = IIf(c1 > c2, LL, IIf(c1 |
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