第二次访问访问C DLL函数时出现Access Violation Exception

Access Violation Exception accessing C DLL function by the second access

我有以下代码:

Generic.cs

class Generic
{
    [DllImport("Generic.dll", EntryPoint = "Consult", CallingConvention = CallingConvention.Winapi)]
    public static extern String Consulta(int NumeroSecao);       
}

main.cs

private void Button_Click(object sender, RoutedEventArgs e)
{
   Random rnd = new Random();
   int random = rnd.Next(9999);
   Stopwatch sw = new Stopwatch();

   sw.Start();
   String teste = Generic.Consult(random);
   sw.Stop();

   TextoLog = "Tempo de Execução:"+sw.Elapsed.Milliseconds+"ms | ConsultarSAT";
   tempoEntreComandos();
}

DLL代码为:

Generic.c

__declspec(dllexport) char* __stdcall Consult(int numeroSessao)
{
    memset(pcReturn,0,sizeof(char) * BUFFER_SIZE);
    HANDLE fileHandle;
    Communicate(&fileHandle, pcReturn,10);
    return pcReturn;
}

当我第二次调用函数Consult时出现问题,收到的消息是Access Violation Exception。这里发生了什么?

解决方法是函数声明中的return的类型,而不是String,必须使用IntPtr.

必须对本机字符串进行封送处理才能在 C# 中使用。您的本机代码实际上并不是 returning 一个字符串对象,而是一个 char 指针。您必须告诉封送拆收器 return 类型是什么,如下所示:

[DllImport("Generic.dll", EntryPoint = "Consult", CallingConvention = CallingConvention.Winapi)]
[return: MarshalAs(UnmanagedType.LPStr)]
public static extern String Consulta(int NumeroSecao);

对 ASCII 字符串使用 LPStr,对 unicode 使用 LPWStr。


实际上,手动编组字符串可能是更好的主意,因为编组器会在转换后立即释放本机字符串。声明将是

[DllImport("Generic.dll", EntryPoint = "Consult", CallingConvention = CallingConvention.Winapi)]
public static extern IntPtr Consulta(int NumeroSecao);

然后您像这样手动将 IntPtr 编组为字符串

String str = Marshal.PtrToStringAnsi(strptr); // for ansi
String str = Marshal.PtrToStringUni(strptr); // for unicode