用vb6,vc6可以画出这样的曲线吗？

c8051fbbs 2011-07-22 04:38:55

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java321java 2011-08-31

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学习了

sachiel8866 2011-08-30

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这个应该没问题的

熊孩子开学喽 2011-08-26

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这种图不难画.

c8051fbbs 2011-08-26

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用mschart

贝隆 2011-08-26

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Option Explicit

Private Type Point

    x As Long

    y As Long

End Type

Private Declare Function CreatePolygonRgn Lib "gdi32" (lpPoint As Any, ByVal nCount As Long, ByVal nPolyFillMode As Long) As Long

Private Declare Function Polygon Lib "gdi32" (ByVal hdc As Long, lpPoint As Any, ByVal nCount As Long) As Long

Private Declare Function FillRgn Lib "gdi32" (ByVal hdc As Long, ByVal hrgn As Long, ByVal hbrush As Long) As Long

Private Declare Function GetStockObject Lib "gdi32" (ByVal nIndex As Long) As Long

Private Declare Function DeleteObject Lib "gdi32" (ByVal hObject As Long) As Long

Const ALTERNATE = 1 ' ALTERNATE and WINDING are

Const WINDING = 2 ' constants for FillMode.

Const BLACKBRUSH = 4 ' Constant for brush type.

Private Const GRAY_BRUSH = 2

Private A(0 To 12) As Point



Private Sub Command1_Click()

    Dim hbrush As Long

    Dim hrgn As Long

    Dim intP As Integer

    

    Polygon Picture1.hdc, A(0), 13

    hbrush = GetStockObject(GRAY_BRUSH)

    hrgn = CreatePolygonRgn(A(0), 13, ALTERNATE)

    If hrgn Then FillRgn Picture1.hdc, hrgn, hbrush

    DeleteObject hrgn

    

    Picture1.DrawWidth = 5

    For intP = 0 To 9

        Picture1.Line (A(intP).x, A(intP).y)-(A(intP + 1).x, A(intP + 1).y), vbBlue

    Next intP

End Sub



Private Sub Form_Load()

    Dim intP As Integer

    For intP = 0 To 10

        A(intP).x = intP * 60

        A(intP).y = 300 - CInt(300 * Rnd)

    Next intP

    A(11).x = 600

    A(11).y = 300

    A(12).x = 0

    A(12).y = 300

    Me.ScaleMode = 3

    Picture1.ScaleMode = 3

    

End Sub