как выйти из аксапты в строго заданном месте?

[michel1971]

определяем ID процесса

static int GetCurrentProcessId()
{
    Dll             kernel32          = new Dll(#KERNELDLL);
    DllFunction     processId         = new DllFunction(kernel32, "GetCurrentProcessId");
    ;
    processId.returns(ExtTypes:: DWord);

    return processId.call();
}

затем убиваем его

static void TerminateProcess(int    _processId)
{
    Dll             kernel32            = new Dll(#KERNELDLL);
    DllFunction     closeHandle         = new DllFunction(kernel32, "CloseHandle");
    DllFunction     terminateProcess    = new DllFunction(kernel32, "TerminateProcess");
    DllFunction     openProcess         = new DllFunction(kernel32, "OpenProcess");

    int             hProcess;

    closeHandle.returns(ExtTypes:: DWORD);
    closeHandle.arg(ExtTypes:: DWORD);

    terminateProcess.returns(ExtTypes:: DWord);
    terminateProcess.arg(ExtTypes:: DWord, ExtTypes:: DWord);

    openProcess.returns(ExtTypes:: DWord);
    openProcess.arg(ExtTypes:: DWord, ExtTypes::DWord, ExtTypes:: DWord);

    hProcess = openProcess.call(1, false, _processId);

    terminateProcess.call(hProcess, 1);

    closeHandle.call(hProcess);
}