在超时中处理GetTickCount()溢出

Question

这是必须在XP上运行的代码，所以没有GetTickCount64，并且在49天后应该正确地处理值的包装。还能改进吗？

// DWORD timeoutMs is a given parameter.

DWORD beginMs = (timeoutMs == INFINITE ? 0 : ::GetTickCount());
DWORD endMs = beginMs + timeoutMs; // unsigned arithmethic is Mod(MAX+1)
DWORD currentMs;

// Create Process, omitted for brevity

while ((waitResult = ::WaitForSingleObject(pi.hProcess, DEFAULT_WAIT_MS)) == WAIT_TIMEOUT)
{
    if (timeoutMs != INFINITE)
    {
        bool timeoutReached = false;
        currentMs = ::GetTickCount();
        if (beginMs <= endMs) // normal case
        {
            if (currentMs > endMs)
            {
                timeoutReached = true;
            }
        }
        else // special case: tick count wrapped around after 49 days uptime
        {
            if (currentMs < beginMs && currentMs > endMs)
            {
                timeoutReached = true;
            }
        }

        if (timeoutReached)
        {
            ::TerminateProcess(pi.hProcess, 0);
            break;
        }
    }
}

Answer 1

无符号减法(和自动mod 2^32)总是给出currentMs - beginMs == elapsedMs，即使是beginMs > currentMs，只要实际运行的时间不溢出滴答计数(它不大于49天)。

因此，您可以用以下代码替换原始代码：

DWORD beginMs = GetTickCount();

// Create Process, omitted for brevity

while ((waitResult = WaitForSingleObject(pi.hProcess, DEFAULT_WAIT_MS)) == WAIT_TIMEOUT) {
    if (timeoutMs != INFINITE) {
        DWORD currentMs = GetTickCount();
        bool timeoutReached = currentMs - beginMs > timeoutMs;
        if (timeoutReached) {
            TerminateProcess(pi.hProcess, 0);
            break;
        }
    }
}

您甚至可以更进一步，消除变量currentMs和timeoutReached：if (GetTickCount() - beginMs > timeoutMs) {...