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Delhi编的一个简单程序：一个窗体，窗体上有一个按键，按鍵后显示对话框“this is only a debug”
unit Unit1;
interface
uses
Windows, Messages, SysUtils, Classes, Graphics, Controls, Forms, Dialogs,
StdCtrls;
type
TForm1 = class(TForm)
Button1: TButton;
procedure Button1Click(Sender: TObject);
private
{ Private declarations }
public
{ Public declarations }
end;
var
Form1: TForm1;
implementation
{$R *.DFM}
procedure TForm1.Button1Click(Sender: TObject);
begin
ShowMessage( 'this is only a debug' );
end;
end.
现在我想直接修改这个编译好的.exe文件，使得按鍵不反应，就是说按键后无条件转移到
ShowMessage函数的后边，我该怎么做？
为什么 bpx shellexecutea 断点设置不成功？怎么做？
unit Unit1;
interface
uses
Windows, Messages, SysUtils, Classes, Graphics, Controls, Forms, Dialogs,
StdCtrls;
type
TForm1 = class(TForm)
Button1: TButton;
procedure Button1Click(Sender: TObject);
private
{ Private declarations }
public
{ Public declarations }
end;
var
Form1: TForm1;
implementation
{$R *.DFM}
procedure TForm1.Button1Click(Sender: TObject);
begin
ShowMessage( 'this is only a debug' );
end;
end.
现在我想直接修改这个编译好的.exe文件，使得按鍵不反应，就是说按键后无条件转移到
ShowMessage函数的后边，我该怎么做？
为什么 bpx shellexecutea 断点设置不成功？怎么做？
