Dll函数调用比普通函数调用快吗？

Question

我正在测试DLL中的导出函数的速度和正常函数.如何在DLL中导出的函数更快？

100000000 function calls in a DLL cost: 0.572682 seconds
100000000 normal function class cost: 2.75258 seconds

这是DLL中的函数:

extern "C" __declspec (dllexport) int example()
{
    return 1;
}

这是正常的函数调用:

int example()
{
    return 1;
}

这是我测试它的方式:

int main()
{
    LARGE_INTEGER frequention;
    LARGE_INTEGER dllCallStart,dllCallStop;
    LARGE_INTEGER normalStart,normalStop;
    int resultCalculation;

    //Initialize the Timer
    ::QueryPerformanceFrequency(&frequention);
    double frequency = frequention.QuadPart;
    double secondsElapsedDll = 0;
    double secondsElapsedNormal = 0;

    //Load the Dll
    HINSTANCE hDll = LoadLibraryA("example.dll");

    if(!hDll)
    {
        cout << "Dll error!" << endl;
        return 0;
    }

    dllFunction = (testFunction)GetProcAddress(hDll, "example");

    if( !dllFunction )
    {
        cout << "Dll function error!" << endl;
        return 0;
    }

    //Dll
    resultCalculation = 0;
    ::QueryPerformanceCounter(&dllCallStart);
    for(int i = 0; i < 100000000; i++)
        resultCalculation += dllFunction();
    ::QueryPerformanceCounter(&dllCallStop);

    Sleep(100);

    //Normal
    resultCalculation = 0;
    ::QueryPerformanceCounter(&normalStart);
    for(int i = 0; i < 100000000; i++)
        resultCalculation += example();
    ::QueryPerformanceCounter(&normalStop);

    //Calculate the result time
    secondsElapsedDll = ((dllCallStop.QuadPart - dllCallStart.QuadPart) / frequency);
    secondsElapsedNormal = ((normalStop.QuadPart - normalStart.QuadPart) / frequency);

    //Output
    cout << "Dll: " << secondsElapsedDll << endl; //0.572682
    cout << "Normal: " << secondsElapsedNormal << endl; //2.75258

    return 0;
}

我只测试函数调用速度,获取地址可以在启动时完成.因此,性能损失并不重要.

Answer 1

对于一个非常小的函数,区别在于函数返回/清除参数的方式.

但是,这不应该产生那么大的差别.我认为编译器意识到你的函数不会对resultCalcuation做任何事情并将其优化掉.尝试使用两个不同的变量,然后打印它们的值.