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Delphi 经典游戏程序设计40例 的学习 例13 各种卸妆效果的处理

本文主要是介绍Delphi 经典游戏程序设计40例 的学习 例13 各种卸妆效果的处理,对大家解决编程问题具有一定的参考价值,需要的程序猿们随着小编来一起学习吧!
unit R13;

interface

uses
  Windows, Messages, SysUtils, Variants, Classes, Graphics, Controls, Forms,
  Dialogs, ExtCtrls, StdCtrls;

type
  TRei13 = class(TForm)
    Button1: TButton;
    Button2: TButton;
    Button3: TButton;
    Button4: TButton;
    Button5: TButton;
    tmr1: TTimer;
    procedure FormCreate(Sender: TObject);
    procedure tmr1Timer(Sender: TObject);
    procedure FormClose(Sender: TObject; var Action: TCloseAction);
    procedure Button1Click(Sender: TObject);
    procedure Button2Click(Sender: TObject);
    procedure Button3Click(Sender: TObject);
    procedure Button4Click(Sender: TObject);
    procedure Button5Click(Sender: TObject);
    procedure FormPaint(Sender: TObject);
  private
    { Private declarations }
    //procedure FmCls(Col:TColor);
    // 在绘制点阵图与画面上绘制图案
    procedure FmDip;
    //procedure DipDt(X1,Y1:Word;Dt:Byte );
    // 在绘制点阵图上依指定的大小及颜色来绘制四方形
    procedure ClsBox(X1,Y1,X2,Y2:Word;Col : TColor );
    //将指定的角色(复合图案)绘制至指定点阵图上
    procedure ChrDi(Xsiz,Ysiz:Byte ;Dpon:Word;X1,Y1:Integer ;Bmap:TBitmap );
    // 将指定的图案绘制至指定点阵图的指定位置上
    procedure PatDi(Pnum:Byte;X1,Y1:Integer;Bmap:TBitmap);
  public
    { Public declarations }
  end;

const
  Yoko = 37;         //横向图案数
  Tate = 27;         // 纵向图案数
  DYoko = Yoko * 16;  // 横向点数 16最小单位?
  Dtate = Tate * 16;
  PtFull = 16;     // 全面显示的(不要重叠显示)的图案数

  MaxSp = 2;      // 本次使用的复合图案总数


var
  Rei13: TRei13;
  LoadBmap,XpatBmap,BpatBmap,BackBmap,MakeBmap : TBitmap;
  // 载入,去除模板,笔刷图案,显示,绘制 点阵图
  // PX ,PY 在PatDi中用, 可以为局部变量。
  //在 TIMER中用,需要
  // p没有用用到,可以去掉

   P,PX,PY,nX,nY,Lp,Cdip : Byte;


  // cdip   被赋值 0,1,但是并没有出现判断的地方,应该可以去掉
  // 测试可以去掉
 // 原来12程序里面的cdip ,  其实是在13程序中被用到的,

   // 用来识别按钮,进入相应的TIMEER,的CASE 功能区  ,255为没有点击按钮用
   Cls:Byte = 255;

  //SC计算零件贴图指针数组的初始设置 ,应该可以放到 FORMCREAT 中
  // n,m TIMEER中做 循环计数器使用,需要退出TMINER 后仍有效,等待下次进入TIMER
  // 所以需要全局变量?
  // 但是 n,在11,13中用,而m 在12,14中用,似乎可以去掉一个。只用一个?
  //测试     确实可以只要一个
  m,{m,}Sc : Word;

  // 需要操作矩形块
  RectL,RectM,RectS,RectD: TRect;

  // 复合图案数组,自定义的数据格式   ,     这个可以整套替换
 { SpSiz: array[0..(MaxSp * 2 - 1)] of Byte = (2,2,31,17) ;
  SpPon :array[0..(MaxSp -1)] of Word;
  SpDat :array[0..(4 + 31  *17 -1)] of Byte =(
  28,29,30,31,
    19,19,19,19,19,19,19,19,19,19,19,19,19,19, 0, 0,
      0,19,19,19,19,19,19,19,19,19,19,19,19,19,19,
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
      0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
      0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
    0,12,12,12,12,12, 0, 0, 0,12,12,12,12,12, 0, 0,
      0,12,12,12,12,12, 0, 0,12,12,12,12,12,12, 0,
    12, 0, 0, 0, 0, 0,12, 0,12, 0, 0, 0, 0, 0,12, 0,
      12, 0, 0, 0, 0, 0,12, 0,12, 0, 0, 0, 0, 0,12,
    12, 0, 0, 0, 0, 0, 0, 0,12, 0, 0, 0, 0, 0,12, 0,
      12, 0, 0, 0, 0, 0,12, 0,12, 0, 0, 0, 0, 0,12,
    12, 0, 0, 0, 0, 0, 0, 0,12, 0, 0, 0, 0, 0,12, 0,
      12, 0, 0, 0, 0, 0,12, 0,12, 0, 0, 0, 0, 0,12,
    12, 0, 0, 0, 0, 0, 0, 0,12, 0, 0, 0, 0, 0,12, 0,
      12, 0, 0, 0, 0, 0,12, 0,12, 0, 0, 0, 0, 0,12,
    12, 0, 0, 0, 0, 0, 0, 0,12, 0, 0, 0, 0, 0,12, 0,
      12, 0, 0, 0, 0, 0,12, 0,12, 0, 0, 0, 0, 0,12,
    12, 0, 0, 0, 0, 0, 0, 0,12, 0, 0, 0, 0, 0,12, 0,
      12, 0, 0, 0, 0, 0,12, 0,12, 0, 0, 0, 0, 0,12,
    12, 0, 0, 0, 0,12,12, 0,12, 0, 0, 0, 0, 0,12, 0,
      12, 0, 0, 0, 0, 0,12, 0,12, 0, 0, 0, 0, 0,12,
    12, 0, 0, 0, 0, 0,12, 0,12, 0, 0, 0, 0, 0,12, 0,
      12, 0, 0, 0, 0, 0,12, 0,12, 0, 0, 0, 0, 0,12,
    12, 0, 0, 0, 0, 0,12, 0,12, 0, 0, 0, 0, 0,12, 0,
      12, 0, 0, 0, 0, 0,12, 0,12, 0, 0, 0, 0, 0,12,
    0,12,12,12,12,12, 0, 0, 0,12,12,12,12,12, 0, 0,
      0,12,12,12,12,12, 0, 0,12,12,12,12,12,12, 0,
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
      0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
      0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
    19,19,19,19,19,19,19,19,19,19,19,19,19,19,19,19,
      19,19,19,19,19,19,19,19,19,19,19,19,19,19,19
  )  ;
   }
  //  复合图案数组
  SpSiz: array[0..(MaxSp * 2 - 1)] of Byte = (2,2, 31,21);
  SpPon: array[0..(MaxSp - 1)] of Word;
  SpDat: array[0..(4 + 31 * 21 - 1)] of Byte = (
    32,33,48,49,
    12,12,12,12,12,12,12,12,12,12,12,12,12,12, 0, 0,
      0,12,12,12,12,12,12,12,12,12,12,12,12,12,12,
    12, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,12, 0,
      0, 0,12, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,12,
    12,12,12,12,12,12,12,12,12,12,12,12,12,12, 0, 0,
      0,12,12,12,12,12,12,12,12,12,12,12,12,12,12,
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
      0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
      0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
    19, 0, 0, 0, 0, 0, 0, 0,19, 0, 0, 0, 0, 0,19, 0,
       0,19,19,19,19,19, 0, 0,19, 0, 0, 0, 0, 0,19,
    19, 0, 0, 0, 0, 0, 0, 0,19, 0, 0, 0, 0, 0,19, 0,
      19, 0, 0, 0, 0, 0,19, 0,19, 0, 0, 0, 0,19, 0,
    19, 0, 0, 0, 0, 0, 0, 0,19, 0, 0, 0, 0, 0,19, 0,
      19, 0, 0, 0, 0, 0, 0, 0,19, 0, 0, 0,19, 0, 0,
    19, 0, 0, 0, 0, 0, 0, 0,19, 0, 0, 0, 0, 0,19, 0,
      19, 0, 0, 0, 0, 0, 0, 0,19, 0, 0,19, 0, 0, 0,
    19, 0, 0, 0, 0, 0, 0, 0,19, 0, 0, 0, 0, 0,19, 0,
      19, 0, 0, 0, 0, 0, 0, 0,19, 0,19, 0, 0, 0, 0,
    19, 0, 0, 0, 0, 0, 0, 0,19, 0, 0, 0, 0, 0,19, 0,
      19, 0, 0, 0, 0, 0, 0, 0,19,19, 0, 0, 0, 0, 0,
    19, 0, 0, 0, 0, 0, 0, 0,19, 0, 0, 0, 0, 0,19, 0,
      19, 0, 0, 0, 0, 0, 0, 0,19, 0,19, 0, 0, 0, 0,
    19, 0, 0, 0, 0, 0, 0, 0,19, 0, 0, 0, 0, 0,19, 0,
      19, 0, 0, 0, 0, 0, 0, 0,19, 0, 0,19, 0, 0, 0,
    19, 0, 0, 0, 0, 0, 0, 0,19, 0, 0, 0, 0, 0,19, 0,
      19, 0, 0, 0, 0, 0, 0, 0,19, 0, 0, 0,19, 0, 0,
    19, 0, 0, 0, 0, 0, 0, 0,19, 0, 0, 0, 0, 0,19, 0,
      19, 0, 0, 0, 0, 0,19, 0,19, 0, 0, 0, 0,19, 0,
    19,19,19,19,19,19,19, 0, 0,19,19,19,19,19, 0, 0,
       0,19,19,19,19,19, 0, 0,19, 0, 0, 0, 0, 0,19,
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
      0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
    12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,
      12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,
    12, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
       2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,12,
    12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,
      12,12,12,12,12,12,12,12,12,12,12,12,12,12,12
  );

implementation

{$R *.dfm}

procedure TRei13.FormCreate(Sender: TObject);
var
  X,Y :Byte;       // 循环用,X,Y 坐标值   绘制BackBmap
  Pn: Word;      //循环用,MAXSP 值,复合图案数组的下标值

begin
  Rei13.Canvas.CopyMode := cmSrcCopy;
  //载入图案
  LoadBmap := TBitmap.Create;
  LoadBmap.LoadFromFile(GetCurrentDir + '\Pat_Sample.bmp');
  //存储去除用点阵图
  XpatBmap := TBitmap.Create;
  XpatBmap.Width := 256;
  XpatBmap.Height := 256;
  // 制作去除用点阵图
  RectL := Rect(0,0,256,256);
  XpatBmap.Canvas.CopyMode := cmSrcCopy;
  XpatBmap.Canvas.CopyRect(RectL,LoadBmap.Canvas,RectL);
  XpatBmap.Canvas.Brush.Color := clBlack;
  XpatBmap.Canvas.BrushCopy(RectL,LoadBmap,RectL,clWhite);
  XpatBmap.Canvas.CopyMode := cmMergePaint;
  XpatBmap.Canvas.CopyRect(RectL,LoadBmap.Canvas,RectL);
  //笔刷点阵图
  BpatBmap := TBitmap.Create;
  BpatBmap.Width := 8;
  BpatBmap.Height := 8;
  //零件贴图指针数组的初始设置
  Sc :=0;
  for Pn :=0 to (MaxSp -1) do
  begin
    SpPon[Pn] := Sc;
    Sc := Sc + SpSiz[Pn * 2] * SpSiz[Pn * 2 + 1]

  end;
  //存储绘制用点阵图
  MakeBmap := TBitmap.Create;
  MakeBmap.Width := DYoko;
  MakeBmap.Height := Dtate;


   //将欲显示的图案绘制到BackBmap
  BackBmap := TBitmap.Create;
  BackBmap.Width := DYoko + 32;
  BackBmap.Height := Dtate + 32;

  for Y := 0 to (Tate -1 )  do
    for X := 0 to (Yoko -1 ) do

    //全部填充 7 图案,该图案是砖块。
    // 填充15 该图案是草地
    PatDi(15,X * 16 + 16,Y * 16 + 16,BackBmap);
    // 画图案0
  ChrDi(SpSiz[0],SpSiz[1],SpPon[0],298,88,BackBmap);
     // 画图案1
  ChrDi(SpSiz[2],SpSiz[3],SpPon[1],66,96,BackBmap);

end;

 { //功能,以指定的颜色将绘制点阵图与画面去除
procedure TRei13.FmCls(Col:TColor);
begin
  RectD :=  Rect(0,0,DYoko,Dtate) ;
  MakeBmap.Canvas.Brush.Color := Col;
  MakeBmap.Canvas.FillRect(RectD);
  Rei12.Canvas.Draw(0,0,MakeBmap);


end;     }


{procedure TRei13.DipDt(X1,Y1:Word;Dt:Byte );
// 从显示画面将指定大小的正方形复制到绘制点阵图
begin
  RectM := Rect(X1 + 16, Y1 + 16,X1 + 16 + Dt,Y1 + 16 + Dt);
  RectD := Rect(X1,Y1,X1 + Dt,Y1 + Dt);
  MakeBmap.Canvas.CopyRect(RectD,BackBmap.Canvas,RectM);

end; }
// 在绘制点阵图与画面上绘制图案
procedure Trei13.FmDip;

begin
  RectM := Rect(16,16,DYoko + 16,Dtate + 16);
  RectD := Rect(0,0,DYoko ,Dtate );
  MakeBmap.Canvas.CopyRect(RectD,BackBmap.Canvas,RectM);
  Rei13.Canvas.Draw(0,0,MakeBmap);

end;
// 在绘制点阵图上依指定的大小及颜色来绘制四方形
procedure trei13.ClsBox(X1,Y1,X2,Y2:Word;Col : TColor);
begin
  RectD := Rect(X1,Y1,X2,Y2);
  MakeBmap.Canvas.Brush.Color := Col;
  MakeBmap.Canvas.FillRect(RectD);

end;

 //将指定的角色(复合图案)绘制至指定点阵图上
procedure TRei13.ChrDi(Xsiz,Ysiz:Byte ;Dpon:Word;X1,Y1:Integer ;Bmap:TBitmap );
var
  CDX,CDY : Byte;
begin
  for CDY := 0 to (Ysiz -1) do
    for CDX := 0 to (Xsiz - 1) do
    begin
      if ( X1 + CDX * 16 >= 0 ) and ( X1 + CDX * 16 <= DYoko + 16 ) and
        (Y1 + CDY * 16 >= 0 ) and (Y1 + CDY * 16 <= DTate + 16 ) then
        PatDi(SpDat[Dpon],X1 + CDX * 16,Y1 + CDY * 16,Bmap );
      Dpon := Dpon + 1;

    end;
end;


 // 将指定的图案绘制至指定点阵图的指定位置上
procedure TRei13.PatDi(Pnum:Byte;X1,Y1:Integer;Bmap:TBitmap);
var
  PX,PY : Byte;
begin
 PX := (Pnum and $F) * 16;    //PX ,PY可以为过程内局部变量码?
 PY := Pnum and $F0;
 RectL := Rect(PX,PY,PX + 16,PY + 16);
 RectD := Rect(X1,Y1,X1 + 16,Y1 + 16);
 if Pnum <> 0 then
 if  Pnum >= PtFull then

 begin
   Bmap.Canvas.CopyMode := cmSrcPaint;
   Bmap.Canvas.CopyRect(RectD,XpatBmap.Canvas,RectL);
   Bmap.Canvas.CopyMode := cmSrcAnd;
   Bmap.Canvas.CopyRect(RectD,LoadBmap.Canvas,RectL);

 end
 else begin
   Bmap.Canvas.CopyMode := cmSrcCopy;
   Bmap.Canvas.CopyRect(RectD,LoadBmap.Canvas,RectL);
 end;

end;

procedure TRei13.tmr1Timer(Sender: TObject);
var

  // 定义局部变量
  X,Y ,Gd:Byte;    // x,y ,坐标循环需要
  n:Word;    //   内部循环计数,
begin

   //化妆显示与画面去除
  case Cls  of
     // 以黑色去除画面并进行初始设置
     // 很有技巧啊,TIMEER是interval 间隔进入。点击第一次进入1..4 初始化一次,
     //再进入 个功能执行完,执行期间用DISP 锁定进入特定11,12,13,14
     // ,      nX,nY ,需要 全局使用?变量M 可以用局部变量,M测试得全局变量
   1..4: begin
     FmDip;
     Cls := Cls + 10;
      m := 0 ;

     nX := 0 ;
     nY := 0 ;

   end;
           //  显示1的处理   ,x,y,x,y 转圈消除  , M 用来计数消除了一圈,也用来定位XY
   11: begin
     MakeBmap.Canvas.CopyMode := cmSrcCopy;
     X := m;
     Y := m;

     if  m < Yoko then

       for X := m to (Yoko * 2 - 2 - m ) do
         ClsBox(X * 8,Y * 8,X * 8 + 8,Y * 8 + 8,clBlack);

       if m < Tate then
         for Y := m  to ( Tate * 2 - 2 - m) do
           ClsBox(X * 8 ,Y * 8 ,X * 8 + 8,Y * 8 + 8,clBlack);

       if m < Yoko then
       for X := (Yoko * 2 - 1 - m) downto (m +1 ) do
        ClsBox(X * 8 ,Y * 8 ,X * 8 + 8,Y * 8 + 8,clBlack);

       if m < Tate then
         for Y := (Tate * 2 - 1 -m) downto (m +1 ) do
         ClsBox(X * 8 ,Y * 8 ,X * 8 + 8,Y * 8 + 8,clBlack);

       m := m + 1;
       //锁定操作
       if (m >= Yoko  ) or (m >= Tate   ) then
         Cls := 0;
       Rei13.Canvas.Draw(0,0,MakeBmap);


   end;
          //倾斜麻点消除效果     ,算了这个效果有点复杂,更多的讲的是技巧,不研究了
   12: begin
      X := nX;
      Y := nY;
      if m < 16 then          // m 用来计数,灰度颜色0-15
        Lp := m
      else
        Lp := 15;

      for Gd := Lp downto 0  do
      begin
        BpatBmap.Canvas.CopyMode := cmSrcCopy;
        RectL := Rect(Gd * 8 + 32,64,Gd * 8 + 40,72);
        RectD := Rect(0,0,8,8);
        BpatBmap.Canvas.CopyRect(RectD,LoadBmap.Canvas,RectL);
        MakeBmap.Canvas.Brush.Bitmap := BpatBmap;    //设置刷子
        MakeBmap.Canvas.CopyMode := cmMergeCopy;

        for n := 0 to (Yoko - 1) do
        begin
          X := (X + 36) mod Yoko ;
          Y := (Y + 26) mod Tate ;
          RectD := Rect(X * 16,Y * 16,X * 16 + 16,Y * 16 + 16);
          MakeBmap.Canvas.CopyRect(RectD,MakeBmap.Canvas,RectD);
        end;
        MakeBmap.Canvas.Brush.Bitmap := nil;

        if Gd = 15 then
        begin
          nX := X;
          nY := Y;

        end;

      end;



      m := m + 1;
      if m = Tate +15 then      //这里为什么要加15? 为什么要加到这么大  27 + 15
      Cls := 0 ;
       Rei13.Canvas.Draw(0,0,MakeBmap);
   end;

   13 : begin
     if m < 16 then
     begin
       //笔刷图案也在载入图案中
       //BpatBmap.Canvas.CopyMode := cmNotSrcCopy;
       BpatBmap.Canvas.CopyMode := cmSrcCopy;
                // 40 = 32 + 8 ,72 = 64 + 8

       RectL := Rect(m * 8 + 32,64,m * 8 + 40,72);
       RectD := Rect(0,0,8,8);
       BpatBmap.Canvas.CopyRect(RectD,LoadBmap.Canvas ,RectL);

       MakeBmap.Canvas.Brush.Bitmap := BpatBmap;
       // CMMERGERCOPY 将原图像和笔刷图案以AND 方式合并并传送出去
       MakeBmap.Canvas.CopyMode := cmMergeCopy;
       RectM := Rect(16,16,DYoko + 16,Dtate + 16);
       RectD := Rect(0,0,DYoko,Dtate);

       //改变的笔刷,还是COPYRECT,
       MakeBmap.Canvas.CopyRect(RectD,BackBmap.Canvas,RectM);
       //释放笔刷图案
       MakeBmap.Canvas.Brush.Bitmap := nil;
       m := m + 1;
     end
     else
       Cls := 0 ;
     Rei13.Canvas.Draw(0,0,MakeBmap);

     end;

  14 : begin
     MakeBmap .Canvas.CopyMode := cmSrcCopy;

       // PN  Y轴上,分2 点  那么可以改 4点?
     for n := 0 to (Tate * 4 -1 ) do
       begin


       RectS := Rect(m * 8,n * 4,DYoko - 8 ,n * 4 + 2);
       RectD := Rect(m * 8 + 8,n * 4,DYoko,n * 4 + 2);
       MakeBmap.Canvas.CopyRect(RectD,makeBmap.Canvas,RectS);
       ClsBox(m * 8,n * 4,m * 8 + 8,n * 4 + 2,clBlack);

       RectS := Rect(8,n * 4 + 2,DYoko - m * 8,n * 4 + 4);
       RectD := Rect(0,n * 4 + 2,DYoko - m * 8-8 ,n * 4 + 4);
       MakeBmap.Canvas.CopyRect(RectD,MakeBmap.Canvas,RectS);
       ClsBox(DYoko - m * 8 - 8 ,n * 4 + 2,DYoko - m * 8,n * 4 + 4,clBlack);



       end;
     // M  移动8 点。 X  轴上
       m  := m + 1;
       if m = Yoko * 2 then
       Cls := 0;
       Rei13.Canvas.Draw(0,0,MakeBmap);

     end;
  255 :
    begin
    FmDip;
    Cls := 0 ;
    Cdip := 1 ;

    end;

   end;
end;

procedure TRei13.FormClose(Sender: TObject; var Action: TCloseAction);
begin
  LoadBmap.Free;
  XpatBmap.Free;
  BpatBmap.Free;
  BackBmap.Free;
  MakeBmap.Free;
end;

procedure TRei13.Button1Click(Sender: TObject);
begin
  //Cdip := 1;
  if Cls = 0 then
     Cls := 1;
end;

procedure TRei13.Button2Click(Sender: TObject);
begin
  //Cdip := 1;
  if Cls = 0 then
    Cls := 2;
end;

procedure TRei13.Button3Click(Sender: TObject);
begin
  //Cdip  := 1;
  if Cls = 0 then
  Cls := 3;
end;

procedure TRei13.Button4Click(Sender: TObject);
begin
  //Cdip := 1;
  if cls = 0 then
  Cls := 4;

end;

procedure TRei13.Button5Click(Sender: TObject);
begin
 // FmCls(clBlack);
  FmDip;
  Cls := 0 ;
  //Cdip := 0 ;
end;
    // form画面重新绘制, 没有redraw ?
procedure TRei13.FormPaint(Sender: TObject);
begin
  Rei13.Canvas.Draw(0,0,MakeBmap);
end;

end.

和例12是相反的功能,结构上一样,所以就在例12的程序上改了一下。

 

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