#include <iostream>
class Foo {
public:
int m_foo;
Foo(int a_foo) : m_foo(a_foo) {}
protected:
bool operator==(const Foo& a) const {
std::cout << "Foo: " << m_foo << " == " << a.m_foo << '\n';
return m_foo == a.m_foo;
}
};
class Bar : public Foo {
public:
int m_bar;
Bar(int a_foo, int a_bar) :
Foo(a_foo),
m_bar(a_bar)
{}
bool operator==(const Bar& a) const {
std::cout << "Bar: " << m_foo << ", " << m_bar << " == " << a.m_foo << ", " << a.m_bar << '\n';
return (const Foo&)*this == (const Foo&)a &&
m_bar == a.m_bar;
}
};
int main() {
Bar a(1, 1);
Bar b(1, 2);
Bar c(2, 2);
std::cout << (a == a) << '\n';
std::cout << (a == b) << '\n';
std::cout << (a == c) << '\n';
return 0;
}
在我的真实代码中,Foo
是一个可以实例化但不应该被允许使用operator==
的类,因此我将其设为protected
。我这样做会遇到编译器错误:
foo.cpp: In member function ‘bool Bar::operator==(const Bar&) const’:
foo.cpp:9:7: error: ‘bool Foo::operator==(const Foo&) const’ is protected
bool operator==(const Foo& a) const {
^
foo.cpp:25:43: error: within this context
return (const Foo&)*this == (const Foo&)a &&
^
为什么不允许这样做?派生类是否应该能够使用protected
方法?
答案 0 :(得分:3)
我无法回答原因(但),但这种替代语法可以满足您的需求:
return Foo::operator==(a) && (m_bar == a.m_bar);