从已删除复制构造函数的类继承

Question

我有一个类base只包含私有默认构造函数和公共删除的复制构造函数,没有别的.

class base {
private:
    base() = default;

public:
    base(const base&) = delete;
};

如果我尝试继承base并创建类的实例,derived如下所示,g ++ 4.8.2不会编译我的代码,但VC++ 2013会编译.

class derived : public base {
private:
    derived() = default;
};

derived x;

那么,它是g ++或VC++ 2013中的一个错误,只是忽略了什么？

这是完整的代码......

class base {
private:
    base() = default;

public:
    base(const base&) = delete;
};

class derived : public base {
private:
    derived() = default;
};

derived x;

int main() { 
}

...和g ++错误消息.

main.cpp:12:5: error: 'constexpr derived::derived()' is private
     derived() = default;
     ^
main.cpp:15:9: error: within this context
 derived x;
         ^
main.cpp: In constructor 'constexpr derived::derived()':
main.cpp:3:5: error: 'constexpr base::base()' is private
     base() = default;
     ^
main.cpp:12:5: error: within this context
     derived() = default;
     ^
main.cpp: At global scope:
main.cpp:15:9: note: synthesized method 'constexpr derived::derived()' first required here
 derived x;
         ^

Answer 1

您误读了错误,它告诉您默认构造函数derived不可访问(是private),因此您无法使用它来创建该类型的对象.现在public在derived级别上创建它将无济于事,因为base构造函数也private因此无法在构造函数中使用derived.

你为什么要那些构造函数private？