方法1:list2.addstrings(list1) 特点是:不会清空list2中原有的数据.

方法2:list2.assign(list1) 特点是:会清空list2中原有的数据(直接替换链表节点的值);

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unit Unit6;

interface

uses
  Winapi.Windows, Winapi.Messages, System.SysUtils, System.Variants, System.Classes, Vcl.Graphics,
  Vcl.Controls, Vcl.Forms, Vcl.Dialogs, Vcl.StdCtrls;

type
  TForm6 = class(TForm)
    Button1: TButton;
    procedure Button1Click(Sender: TObject);
  private
    { Private declarations }
  public
    { Public declarations }
  end;

var
  Form6: TForm6;

implementation

{$R *.dfm}

procedure TForm6.Button1Click(Sender: TObject);
var
  list1, list2, list3: TStringList;
begin
  list1 := TStringList.Create;
  list2 := TStringList.Create;
  list3 := TStringList.Create;
  try
    list1.Add('a');
    list1.Add('b');
    list1.Add('c');

    list2.Add('1');
    list2.AddStrings(list1);

    list3.Add('2');
    list3.Assign(list1);


    ShowMessage(list1.Text);
    ShowMessage('留意这里1不会被清空' + sLineBreak + list2.Text);
    ShowMessage('2会被清空' + sLineBreak + list3.Text);

    list1.Clear;
    ShowMessage('由于是堆中数据被复制一份,所以清空list1后,不会影响list2,list3');

    ShowMessage('已被清空' + sLineBreak + list1.Text);
    ShowMessage('不变' + sLineBreak + list2.Text);
    ShowMessage('不变' + sLineBreak + list3.Text);
  finally
    list1.Free;
    list2.Free;
    list3.Free;
  end;
end;

end.

 

posted on 2017-08-16 11:37  del88  阅读(382)  评论(0编辑  收藏  举报