Delphi“E2064 左侧无法分配给”将项目从2009升级到XE时出现错误

发布于 2024-10-27 22:04:07 字数 945 浏览 4 评论 0原文

我读过这个问题，其中相同问题已讨论，无论如何，我能够在 Delphi 2009 中做到这一点，但当我升级到 XE 时，这是不可能的。

我在这里粘贴一个简单的虚拟示例：它在 2009 年编译并在 XE 上给出 E2064...为什么？是否可以将 XE 设置为像 2009 一样？或者我应该寻求解决方法？

unit Unit2;

interface

uses
  Windows, Messages, SysUtils, Variants, Classes, Graphics, Controls, Forms,
  Dialogs, StdCtrls;
type
  TTestRecord = record
    FirstItem  : Integer;
    SecondItem  : Integer;
  end;
  TForm2 = class(TForm)
    procedure AssignValues;
  private
    FTestRecord :TTestRecord;
  public
    property TestRecord : TTestRecord read FTestRecord write FTestRecord;
  end;

var
  Form2: TForm2;

implementation

{$R *.dfm}

procedure TForm2.AssignValues;
begin
with TestRecord do
     begin
       FirstItem := 14; // this gives error in XE but not in 2009
       SecondItem := 15;
     end;
end;

end.

原文

I read this question in which the same problem is discussed, anyway i was able to do this in Delphi 2009 and this was not possible as i upgraded to XE.

I paste here a simple dummy example: this compiles on 2009 and gives E2064 on XE... Why? Is it possible to setup XE to behave like 2009? Or should I go for a workaround?

unit Unit2;

interface

uses
  Windows, Messages, SysUtils, Variants, Classes, Graphics, Controls, Forms,
  Dialogs, StdCtrls;
type
  TTestRecord = record
    FirstItem  : Integer;
    SecondItem  : Integer;
  end;
  TForm2 = class(TForm)
    procedure AssignValues;
  private
    FTestRecord :TTestRecord;
  public
    property TestRecord : TTestRecord read FTestRecord write FTestRecord;
  end;

var
  Form2: TForm2;

implementation

{$R *.dfm}

procedure TForm2.AssignValues;
begin
with TestRecord do
     begin
       FirstItem := 14; // this gives error in XE but not in 2009
       SecondItem := 15;
     end;
end;

end.

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烙印 2024-11-03 22:04:07

D2010 编译器只是比以前的版本更严格。在以前的版本中，编译器不会抱怨，但结果通常不会像您预期的那样，因为它作用于临时变量，因此您的更改将在方法结束时消失。

您链接到的问题的答案提供了更好的解释，并提供了可供选择的解决方案（或解决方法）。

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々眼睛长脚气 2024-11-03 22:04:07

好吧好吧，对不起，我不应该做出非技术性的内容……

现在，我们可以修改代码如下，就可以正常工作了：

type
  PTestRecord = ^TTestRecord;
  TTestRecord = record
    FirstItem: Integer;
    SecondItem: Integer;
  end;

  TForm2 = class(TForm)
  private
    { Private declarations }
    FTestRecord: TTestRecord;
    procedure AssignValues;
  public
    { Public declarations }
    property TestRecord: TTestRecord read FTestRecord write FTestRecord;
  end;

var
  Form2: TForm2;

implementation

{$R *.dfm}

procedure TForm2.AssignValues;
begin
  with PTestRecord(@TestRecord)^ do
  begin
    FirstItem := 14; // it works fine.
    SecondItem := 15;
  end;
end;

OK, OK, I am sorry, I shouldn't have made non-technical content...

Now, we can modify the code as follows, and it works fine：

type
  PTestRecord = ^TTestRecord;
  TTestRecord = record
    FirstItem: Integer;
    SecondItem: Integer;
  end;

  TForm2 = class(TForm)
  private
    { Private declarations }
    FTestRecord: TTestRecord;
    procedure AssignValues;
  public
    { Public declarations }
    property TestRecord: TTestRecord read FTestRecord write FTestRecord;
  end;

var
  Form2: TForm2;

implementation

{$R *.dfm}

procedure TForm2.AssignValues;
begin
  with PTestRecord(@TestRecord)^ do
  begin
    FirstItem := 14; // it works fine.
    SecondItem := 15;
  end;
end;

回复收藏 0 原文

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