在TThread
实例创建和启动之间,主线程将继续执行代码。如果主线程中的代码依赖于有问题的线程完全启动并运行,它必须以某种方式等待线程Execute
方法实际启动。
考虑以下代码:
const
WM_MY_ACTION = WM_APP + 10;
type
TWndThread = class(TThread)
protected
fWndHandle: THandle;
IsRunning: boolean;
procedure WndProc(var Msg: TMessage);
procedure Execute; override;
public
Test: integer;
procedure AfterConstruction; override;
procedure DoAction;
end;
procedure TWndThread.AfterConstruction;
begin
inherited;
while not IsRunning do Sleep(100); // wait for thread start up
end;
procedure TWndThread.Execute;
var
Msg: TMsg;
begin
fWndHandle := AllocateHWnd(WndProc);
IsRunning := true;
try
while not Terminated do
begin
if MsgWaitForMultipleObjects(0, nil^, False, 1000, QS_ALLINPUT) = WAIT_OBJECT_0 then
begin
while PeekMessage(Msg, 0, 0, 0, PM_REMOVE) do
begin
TranslateMessage(Msg);
DispatchMessage(Msg);
end;
end;
end;
finally
DeallocateHWnd(fWndHandle);
end;
end;
procedure TWndThread.WndProc(var Msg: TMessage);
begin
case Msg.Msg of
WM_MY_ACTION:
begin
inc(Test);
end;
else Msg.Result := DefWindowProc(fWndHandle, Msg.Msg, Msg.WParam, Msg.LParam);
end;
end;
procedure TWndThread.DoAction;
begin
PostMessage(fWndHandle, WM_MY_ACTION, 0, 0);
end;
var
t: TWndThread;
begin
t := TWndThread.Create;
t.DoAction;
t.Terminate;
end;
如果没有等待IsRunning
标志的循环,DoAction
将无法成功将消息发布到包含的窗口句柄,因为它尚未创建。基本上不会触发inc(Test)
里面的。WndProc
有没有更好的方法来等待线程启动并在Execute
方法内部完成必要的初始化,或者这个解决方案是否尽可能好?
注意:我知道这不是线程安全的AllocateHWnd
,DeallocateHWnd
不应该像上面的例子一样在生产代码中使用。