C ++与继承类的奇怪性能

时间:2015-12-26 02:06:07

标签: c++ performance

据我所知,继承层次结构中的C ++类中的虚函数调用应该比不从任何基类继承的等效类(即自包含类)慢一点。我决定写一个小测试程序来看看性能的差异。

我有一个由3个类组成的继承层次结构:Shape,Rectangle,Quadrilateral。我有一个名为BaseQuadrilateral的类,它不会从任何东西继承,并且做与Quadrilateral类相同的事情。每个类有两种方法:surfaceArea()和volume()。我在每个类上运行一个单独的基准测试,并记录运行10,000,000个对象所需的时间。我预计四边形课程会花费更长的时间。相反,Quadrilateral类(继承自Rectangle)的运行速度比BaseQuadrilateral快。我不明白为什么会这样。

Test Results:
Running Dynamic Dispatch Test

Quadrilateral Runtime: 2840264 Ticks, 2 Seconds.
BaseQuadrilateral Runtime: 21179219 Ticks, 21 Seconds.

有人可以向我解释幕后发生的事情会使继承的代码更快,在什么情况下,继承的代码运行速度会慢于非继承的代码。

谢谢

class Shape
{
public:
    virtual double surfaceArea() = 0;
    virtual double Volume() = 0;
};

class Rectangle : public Shape
{
public:
    //Constructors
    Rectangle();
    Rectangle(double, double);
    Rectangle(const Rectangle&);

    Rectangle& operator=(const Rectangle&);
    double Area();

    //Override Shape base class methods
    double surfaceArea();
    double Volume();
protected:
    double length;
    double width;
};

class Quadrilateral : public Rectangle
{
public:
    //Constructors
    Quadrilateral();
    Quadrilateral(double, double, double);
    Quadrilateral(const Quadrilateral&);

    Quadrilateral& operator=(const Quadrilateral&);

    //Overloaded Square base class
    double surfaceArea();
    double Volume();
protected:
    double height;
};

class BaseQuadrilateral
{
public:
    //Constructors
    BaseQuadrilateral();
    BaseQuadrilateral(double, double, double);
    BaseQuadrilateral(const BaseQuadrilateral&);

    BaseQuadrilateral& operator=(const BaseQuadrilateral&);
    double surfaceArea();
    double Volume();

protected:
    double length;
    double width;
    double height;
};

void test2()
{
    clock_t qTimer, bqTimer;
    Quadrilateral* quadrilaterals;
    BaseQuadrilateral* baseQuadrilaterals, baseQuadrilateral;
    Shape* shape;
    double* answers1, *answers2;
    srand((unsigned int)time(NULL));

    cout << "Running Dynamic Dispatch Test\n" << endl;

    quadrilaterals = new Quadrilateral[ARRAY_SIZE];
    baseQuadrilaterals = new BaseQuadrilateral[ARRAY_SIZE];
    answers1 = new double[ARRAY_SIZE];
    answers2 = new double[ARRAY_SIZE];

    //Initialization
    for (int i = 0; i < ARRAY_SIZE; i++)
    {
        double length = (double)(rand() % 100);
        double width = (double)(rand() % 100);
        double height = (double)(rand() % 100);

        quadrilaterals[i] = Quadrilateral(length, width, height);
        baseQuadrilaterals[i] = BaseQuadrilateral(length, width, height);
    }

    //Test Shape
    qTimer = clock();

    for (int i = 0; i < ARRAY_SIZE; i++)
    {
        shape = &quadrilaterals[i];
        answers1[i] = shape->Volume();
    }

    qTimer = clock() - qTimer;

    //Test BaseQuadrilateral
    bqTimer = clock();

    for (int i = 0; i < ARRAY_SIZE; i++)
    {
        baseQuadrilateral = baseQuadrilaterals[i];
        answers2[i] = baseQuadrilateral.Volume();
    }

    bqTimer = clock() - qTimer;

    for (int i = 0; i < ARRAY_SIZE; i++)
    {
        if (answers1[i] != answers2[i])
        {
            cout << "Incorrect answer found at i=" << i << ". answers1: " << answers1[i] << " answers2: " << answers2[i] << endl;
            break;
        }
    }

    //Print Results
    cout << "Quadrilateral Runtime: " << qTimer << " Ticks, " << qTimer / CLOCKS_PER_SEC << " Seconds." << endl;
    cout << "BaseQuadrilateral Runtime: " << bqTimer << " Ticks, " << bqTimer / CLOCKS_PER_SEC << " Seconds." << endl;
}

2 个答案:

答案 0 :(得分:3)

正如我在评论中写的那样,你对这一行有一点问题:

bqTimer = clock() - qTimer;

。您似乎希望qTimerbqTimer。然而,这并不能解释您观察到的差异。为此,您应该仔细查看两个测试循环,特别是它们的不同之处。

在第一种情况下,您正在变量Quadrilateral中记录指向shape的指针,然后通过指针间接调用Volume()方法:

for (int i = 0; i < ARRAY_SIZE; i++)
{
    shape = &quadrilaterals[i];
    answers1[i] = shape->Volume();
}

在第二种情况下,您正在复制整个BaseQuadrilateral对象,然后调用副本的Volume()方法:

for (int i = 0; i < ARRAY_SIZE; i++)
{
    baseQuadrilateral = baseQuadrilaterals[i];
    answers2[i] = baseQuadrilateral.Volume();
}

复制对象比获取对象要昂贵得多。实际上,在这种特殊情况下,甚至可以完全优化地址计算。我建议避免使用中间件,并且在两种情况下直接在数组元素上调用方法:

    answers1[i] = quadrilaterals[i].Volume();

    answers2[i] = baseQuadrilaterals[i].Volume();

答案 1 :(得分:0)

该行

bqTimer = clock() - qTimer;

bqTimer = clock() - bqTimer;