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  • Delphi 中 "位" 的使用(2) 集合


    集合中的每个元素其实只占一个二进制位, 不足 8 个元素的集合只需要 1 个字节.

    先观察集合的大小:

    Type
      TSet1 = set of (a1,a2,a3,a4,a5,a6,a7,a8);    {刚好对应一个字节的 8 个位}
      TSet2 = set of (b1,b2,b3);                   {只用一个字节中的 3 个位, 也要占一个字节}
      TSet3 = set of (c1,c2,c3,c4,c5,c6,c7,c8,c9); {需要 9 个位, 一个字节容不下了}
      TSet4 = set of Char;
    
    procedure TForm1.Button1Click(Sender: TObject);
    begin
      ShowMessageFmt('%d, %d, %d, %d', [SizeOf(TSet1), //1
                                        SizeOf(TSet2), //1
                                        SizeOf(TSet3), //2
                                        SizeOf(TSet4)  //32
                                       ]);
    end;
    

    洞察集合的二进制表示:

    {查看二进制的函数}
    function ToBin(p: PByteArray; b: Integer): string;
    var
      i,j: Integer;
    begin
      Result := StringOfChar('0', b * 8);
      for i := 0 to b - 1 do for j := 0 to 7 do
        if Odd(p^[b-1-i] shr j) then Result[i*8 + 8 - j] := '1';
    end;
    
    procedure TForm1.Button1Click(Sender: TObject);
    var
      set1: set of (A,B,C,D,E,F,G,H);
    begin
      set1 := [];
      ShowMessage(ToBin(@set1, SizeOf(set1))); //00000000
      set1 := [A,B,C,D,E,F,G,H];
      ShowMessage(ToBin(@set1, SizeOf(set1))); //11111111
      set1 := [A,B,C];
      ShowMessage(ToBin(@set1, SizeOf(set1))); //00000111
      set1 := [A,B,C,H];
      ShowMessage(ToBin(@set1, SizeOf(set1))); //10000111
    end;
    

    甚至可以把集合看成一个数字:

    procedure TForm1.Button1Click(Sender: TObject);
    type
      TSet = set of (A,B,C,D,E,F,G,H);
    var
      s1,s2,s3,s4: TSet;
    begin
      s1 := [];
      s2 := [A,B,C,D,E,F,G,H];
      s3 := [A,B,C];
      s4 := [A,B,C,H];
      ShowMessage(IntToStr(Byte(s1))); //0
      ShowMessage(IntToStr(Byte(s2))); //255
      ShowMessage(IntToStr(Byte(s3))); //7
      ShowMessage(IntToStr(Byte(s4))); //135
    end;
    

    用集合的方式重新做前一次的例子(窗体设计与测试效果同前):

    var set1: set of 0..7; {准备用自定义的集合变量 set1 储存下面的 8 种状态}
    
    procedure TForm1.FormCreate(Sender: TObject);
    begin
      CheckListBox1.Items.CommaText := 'A,B,C,D,E,F,G,H';
      Button1.Caption := '保存状态';
      Button2.Caption := '恢复状态';
      Button3.Caption := '全选';
      Button4.Caption := '全不选';
      Button1.Tag := 1;
      Button2.Tag := 2;
      Button3.Tag := 3;
      Button4.Tag := 4;
      Button2.OnClick := Button1.OnClick;
      Button3.OnClick := Button1.OnClick;
      Button4.OnClick := Button1.OnClick;
    end;
    
    procedure TForm1.Button1Click(Sender: TObject);
    var
      i: Integer;
    begin
      if TButton(Sender).Tag = 1 then set1 := [];
      for i := 0 to CheckListBox1.Count - 1 do
        case TButton(Sender).Tag of
          1: if CheckListBox1.Checked[i] then Include(set1, i);
          2: CheckListBox1.Checked[i] := i in set1;
          3: CheckListBox1.Checked[i] := True;
          4: CheckListBox1.Checked[i] := False;
        end;
    end;
    

    实例观察 TFontStyles 集合:

    {查看二进制的函数}
    function ToBin(p: PByteArray; b: Integer): string;
    var
      i,j: Integer;
    begin
      Result := StringOfChar('0', b * 8);
      for i := 0 to b - 1 do for j := 0 to 7 do
        if Odd(p^[b-1-i] shr j) then Result[i*8 + 8 - j] := '1';
    end;
    
    procedure TForm1.Button1Click(Sender: TObject);
    var
      fs: TFontStyles;
    begin
      Font.Style := [fsBold, fsItalic, fsUnderline];
      fs := Font.Style;
      Text := ToBin(@fs, SizeOf(fs));
    end;
    
    procedure TForm1.Button2Click(Sender: TObject);
    var
      fs: TFontStyles;
    begin
      Font.Style := [];
      fs := Font.Style;
      Text := ToBin(@fs, SizeOf(fs));
    end;
    

    TFontStyles 集合的测试效果图:



    接下来学习 TBits 类; 对 "位" 的操作 TBits 应该是最直观的.

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  • 原文地址:https://www.cnblogs.com/del/p/1730116.html
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