c – 覆盖虚拟功能和隐藏非虚拟功能有什么区别？

给定以下代码片段,函数调用有何不同？什么是隐藏功能？什么是功能重写？它们如何与函数重载相关？两者有什么区别？我在一个地方找不到这些的好描述,所以我在这里问我所以我可以巩固这些信息.

class Parent {
  public:
    void doA() { cout << "doA in Parent" << endl; }
    virtual void doB() { cout << "doB in Parent" << endl; }
};

class Child : public Parent {
  public:
    void doA() { cout << "doA in Child" << endl; }
    void doB() { cout << "doB in Child" << endl; }
};

Parent* p1 = new Parent();
Parent* p2 = new Child();
Child* cp = new Child();

void testStuff() {
  p1->doA();
  p2->doA();
  cp->doA();

  p1->doB();
  p2->doB();
  cp->doB();
}

解决方法:

什么是隐藏功能？

……是隐藏名称的一种形式.一个简单的例子：

void foo(int);
namespace X
{
    void foo();

    void bar()
    {
        foo(42); // will not find `::foo`
        // because `X::foo` hides it
    }
}

这也适用于基类中的名称查找：

class Base
{
public:
    void foo(int);
};

class Derived : public Base
{
public:
    void foo();
    void bar()
    {
        foo(42); // will not find `Base::foo`
        // because `Derived::foo` hides it
    }
};

什么是功能重写？

这与虚函数的概念有关. [class.virtual] / 2

If a virtual member function vf is declared in a class Base and in a class Derived, derived directly or indirectly from Base, a member function vf with the same name, parameter-type-list, cv-qualification, and ref-qualifier (or absence of same) as Base::vf is declared, then Derived::vf is also virtual (whether or not it is so declared) and it overrides Base::vf.

class Base
{
private:
    virtual void vf(int) const &&;
    virtual void vf2(int);
    virtual Base* vf3(int);
};

class Derived : public Base
{
public: // accessibility doesn't matter!
    void vf(int) const &&; // overrides `Base::vf(int) const &&`
    void vf2(/*int*/);     // does NOT override `Base::vf2`
    Derived* vf3(int);     // DOES override `Base::vf3` (covariant return type)
};

调用虚函数时,最终的覆盖变得相关：[class.virtual] / 2

A virtual member function C::vf of a class object S is a final overrider unless the most derived class of which S is a base class subobject (if any) declares or inherits another member function that overrides vf.

即如果你有一个S类型的对象,那么最终的覆盖是你在遍历S的类层次结构回到它的基类时看到的第一个覆盖.重要的是,函数调用表达式的动态类型用于确定最终的覆盖：

Base* p = new Derived;
p -> vf();    // dynamic type of `*p` is `Derived`

Base& b = *p;
b  . vf();    // dynamic type of `b` is `Derived`

覆盖和隐藏有什么区别？

实质上,基类中的函数总是被派生类中的同名函数隐藏;无论派生类中的函数是否覆盖基类的虚函数：

class Base
{
private:
    virtual void vf(int);
    virtual void vf2(int);
};

class Derived : public Base
{
public:
    void vf();     // doesn't override, but hides `Base::vf(int)`
    void vf2(int); // overrides and hides `Base::vf2(int)`
};

要查找函数名称,请使用表达式的静态类型：

Derived d;
d.vf(42);   // `vf` is found as `Derived::vf()`, this call is ill-formed
            // (too many arguments)

它们如何与函数重载相关？

由于“函数隐藏”是一种名称隐藏形式,如果隐藏了函数的名称,则所有重载都会受到影响：

class Base
{
private:
    virtual void vf(int);
    virtual void vf(double);
};

class Derived : public Base
{
public:
    void vf();     // hides `Base::vf(int)` and `Base::vf(double)`
};

对于函数重写,只覆盖具有相同参数的基类中的函数;你当然可以重载一个虚函数：

class Base
{
private:
    virtual void vf(int);
    virtual void vf(double);
    void vf(char);  // will be hidden by overrides in a derived class
};

class Derived : public Base
{
public:
    void vf(int);    // overrides `Base::vf(int)`
    void vf(double); // overrides `Base::vf(double)`
};

标签：method-hiding,c,inheritance,virtual-functions,overriding
来源： https://codeday.me/bug/20190918/1811862.html