如果在结构中再拿出 4 个字节的地址空间指向上一个节点, 就成了双向链表了.

本例效果图:


unit Unit1;

interface

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

type
  TForm1 = class(TForm)
    Button1: TButton;
    Button2: TButton;
    procedure FormCreate(Sender: TObject);
    procedure Button1Click(Sender: TObject);
    procedure Button2Click(Sender: TObject);
  end;

var
  Form1: TForm1;

implementation

{$R *.dfm}

type
  PMyRec = ^MyRec;
  MyRec = record
    Name: string[8];
    Age : Word;
    Prev,Next: PMyRec; {上一个与下一个的结构指针}
  end;

var
  CurrentNode: PMyRec;

{建立双向链表}
procedure TForm1.FormCreate(Sender: TObject);
var
  PrevNode: PMyRec;
  i: Integer;
begin
  PrevNode := nil;
  for i := 65 to 71 do  {建七个节点, 准备用 AA, BB, CC, DD, EE, FF, GG 来指定 MyRec.Name }
  begin
    New(CurrentNode);
    CurrentNode.Name := Chr(i) + Chr(i);
    CurrentNode.Age := i;
    CurrentNode.Prev := PrevNode;
    if Assigned(PrevNode) then PrevNode.Next := CurrentNode;
    PrevNode := CurrentNode;
  end;

  Button1.Caption := '上一个';
  Button2.Caption := '下一个';
end;

{上一个}
procedure TForm1.Button1Click(Sender: TObject);
begin
  if Assigned(CurrentNode.Prev) then
  begin
    CurrentNode := CurrentNode.Prev;
    Text := CurrentNode.Name + #44 + IntToStr(CurrentNode.Age);
  end else Text := '这是第一个节点';
end;

{下一个}
procedure TForm1.Button2Click(Sender: TObject);
begin
  if Assigned(CurrentNode.Next) then
  begin
    CurrentNode := CurrentNode.Next;
    Text := CurrentNode.Name + #44 + IntToStr(CurrentNode.Age);
  end else Text := '这是最后一个节点';
end;

end.

posted on 2008-03-19 13:32  万一  阅读(4834)  评论(8编辑  收藏  举报