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.

posted on 2010-05-07 22:57  万一  阅读(2866)  评论(0编辑  收藏  举报