只允許使用信號量的應用程序的3個實例

Question

我想實現一個簡單的例程使用信號量，這將允許我只運行應用程序的3個實例。我可以用3個互斥但這並不是一個很好的辦法我想這至今

var 
    hSem:THandle; 
begin 
    hSem := CreateSemaphore(nil,3,3,'MySemp3'); 
    if hSem = 0 then 
    begin 
    ShowMessage('Application can be run only 3 times at once'); 
    Halt(1); 
    end; 

我怎樣才能做到這正常嗎？

Answer 1

請務必確保您釋放信號量，因爲如果您的應用程序死亡，這不會自動完成。

program Project72; 

{$APPTYPE CONSOLE} 

uses 
    Windows, 
    SysUtils; 

var 
    hSem: THandle; 

begin 
    try 
    hSem := CreateSemaphore(nil, 3, 3, 'C15F8F92-2620-4F3C-B8EA-A27285F870DC/myApp'); 
    Win32Check(hSem <> 0); 
    try 
     if WaitForSingleObject(hSem, 0) <> WAIT_OBJECT_0 then 
     Writeln('Cannot execute, 3 instances already running') 
     else begin 
     try 
      // place your code here 
      Writeln('Running, press Enter to stop ...'); 
      Readln; 
     finally ReleaseSemaphore(hSem, 1, nil); end; 
     end; 
    finally CloseHandle(hSem); end; 
    except 
    on E: Exception do 
     Writeln(E.ClassName, ': ', E.Message); 
    end; 
end. 

Answer 2

您可以使用JCL的TJclAppInstances。

Answer 3

1. 你一定要嚐嚐，看它是否被創建
2. 必須使用等待函數之一，看看你可以得到一個數
3. 最後，你必須釋放鎖&處理，使其能夠工作在下次用戶關閉和打開應用

乾杯

var 
    hSem: THandle; 
begin 
    hSem := OpenSemaphore(nil, SEMAPHORE_ALL_ACCESS, True, 'MySemp3'); 
    if hSem = 0 then 
    hSem := CreateSemaphore(nil, 3, 3,'MySemp3'); 

    if hSem = 0 then 
    begin 
    ShowMessage('... show the error'); 
    Halt(1); 
    Exit;  
    end; 

    if WaitForSingleObject(hSem, 0) <> WAIT_OBJECT_0 then 
    begin 
    CloseHandle(hSem); 
    ShowMessage('Application can be runed only 3 times at once'); 
    Halt(1); 
    Exit; 
    end; 

    try 
    your application.run codes.. 

    finally 
    ReleaseSemaphore(hSem, 1, nil); 
    CloseHandle(hSem); 
    end;