C ++继承和函数覆盖 [英] C++ inheritance and function overriding
问题描述
#include< iostream>
using std :: cout;
using std :: endl;
class A {};
class B {};
class X
{
public:
void spray(A&)
{
cout< A类<< endl;
}
};
class Y:public X
{
public:
void spray(B&)
{
cout< B类<< endl;
}
};
int main()
{
A a;
B b;
Y y;
y.spray(a);
y.spray(b);
return 0;
}
GCC throws
错误:没有匹配的函数调用`Y :: spray(A&)'
注意:候选项是:void Y :: spray(B&)
用于描述这个术语的术语是隐藏,而不是 。默认情况下,派生类的成员将使任何具有相同名称的基类的成员无法访问,无论它们是否具有相同的签名。如果要访问基类成员,可以使用使用
声明将它们拉入派生类。在这种情况下,使用X添加以下到 class Y
:
::喷雾;
In C++, will a member function of a base class be overridden by its derived class function of the same name, even if its prototype (parameters' count, type and constness) is different? I guess this a silly question, since many websites says that the function prototype should be the same for that to happen; but why doesn't the below code compile? It's a very simple case of inheritance, I believe.
#include <iostream>
using std::cout;
using std::endl;
class A {};
class B {};
class X
{
public:
void spray(A&)
{
cout << "Class A" << endl;
}
};
class Y : public X
{
public:
void spray(B&)
{
cout << "Class B" << endl;
}
};
int main()
{
A a;
B b;
Y y;
y.spray(a);
y.spray(b);
return 0;
}
GCC throws
error: no matching function for call to `Y::spray(A&)'
note: candidates are: void Y::spray(B&)
The term used to describe this is "hiding", rather than "overriding". A member of a derived class will, by default, make any members of base classes with the same name inaccessible, whether or not they have the same signature. If you want to access the base class members, you can pull them into the derived class with a using
declaration. In this case, add the following to class Y
:
using X::spray;
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