虚拟函数查找的规则是什么? [英] What are the rules for virtual function lookup?

问题描述

#include <iostream>
class base
{
    public:
    virtual void print (int a)
    {   
        std::cout << "a: " << a << " base\n";
    }   
    virtual void print (int a, int b)
    {   
        std::cout << "base\n";
    }   
};

class derived : public base
{
    public:
    virtual void print (double d)
    {   
        std::cout << "derived\n";
    }   
};

int main ()
{
    int i = 10; 
    double d = 10000.0;
    base *b = new derived (); 
    b->print (i, i); 
    b->print (d);

    return 0;
}

此函数的输出为:

base
a: 10000 base

• 为什么b->print (d)不调用派生类的实现，并且 对'd'执行静态强制转换以提供与基类的匹配 实施?
• 在查找虚拟函数期间在此处应用什么规则?
• Why b->print (d) don't invoke the derived class implementation and performs static cast on 'd' to provide a match with base class implementation ?
• What rule is applied here during virtual function lookup ?

推荐答案

derived::print不会覆盖base中的任何成员函数.它被声明为具有单个double类型的参数，但是在base中被命名为print的两个虚拟成员函数被声明为具有一个和两个类型为int的参数.

derived::print does not override any member function in base. It is declared as having a single parameter of type double but the two virtual member functions named print in base are declared as having one and two parameters of type int.

使用b->print(d)时，在重载解析期间仅考虑base中的成员函数，因此仅考虑void base::print(int)和void base::print(int, int).找不到void derived::print(double)，因为编译器不知道b指向derived对象.

When you use b->print(d), only member functions in base are considered during overload resolution, so only void base::print(int) and void base::print(int, int) are considered. void derived::print(double) can't be found because the compiler has no idea that b points to a derived object.

如果derived要覆盖在base中声明为虚拟成员函数的两个print函数之一，则该覆盖将在运行时被调用.

If derived were to override one of the two print functions declared as virtual member functions in base, then that override would be called at runtime.

(有点相关的注释，derived::print隐藏了两个base::print成员函数，因此，如果您尝试使用基类print函数之一，例如derived().print(1, 1)，它将失败.您将需要使用using声明，以使这些成员函数在名称查找期间可用.)

(On a somewhat related note, derived::print hides the two base::print member functions, so if you were to try to use one of the base class print functions, e.g., derived().print(1, 1), it would fail. You would need to use a using declaration to make those member functions available during name lookup.)

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