Question

在一次采访中有人问我这个问题。我的答案是（3和3.6）（错误）。请解释一下我的理解是错的

我的想法是指针 bd 将指向派生类vtable的_vptr。

派生类的Vtable将包含2个函数

double func(double) // ----->points to Derived::func()
int func(int)       // ----->points to Base::func()

因此

bd->func(2)   // will call Base::func() i.e int func(int)
bd->func(2.3) // will call Derived::func() i.e double func(double)

请说明我的理解是错误的。另外，请说明Base::func()不是virtual时的情况。在那种情况下，将不会有vtable吗？函数调用将如何解决？

#include <iostream>    
using namespace std;

class Base
{
private:
    /* data */
public:
    Base(/* args */){};
    ~Base(){};

    //int func(int i)     getting same answer regardless of virtual
    virtual int func(int i)
    {
        cout << "Base func()" << endl;
        return i+1;
    }
};

class Derived : public Base
{
public:
    Derived(/* args */){};
    ~Derived(){};

    double func(double d)
    {
        cout << "Derived func()" << endl;
        return d+1.3;
    }
};

int main() {
    Base* bd = new Derived();
    cout << bd->func(2) << endl;
    cout << bd->func(2.3) << endl;
    return 0;
}

预期输出：

Base func()
3
Derived func()
3.6

Actual output:
Base func()
3
Base func()
3

Answer 1

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Base::func(int i)根本不在此代码中发挥作用。

Derived::func(double d)中的func是virtual并不重要，因为Base中的func并不Derived（签名不匹配）。将override关键字添加到override将使编译器错误立即变得清楚。

具有相同名称但参数和返回类型不同的虚函数

1 个答案: