operator =和在C ++中不继承的函数? [英] operator= and functions that are not inherited in C++?
问题描述
直到我刚才做的测试,我相信只有构造函数不是在C ++中继承的。但显然,赋值 operator =
不是太...
Until a test I've just made, I believed that only Constructors were not inherited in C++. But apparently, the assignment operator=
is not too...
- 什么是原因是什么?
- 是否有任何解决方法来继承赋值运算符?
-
operator + =
,operator - =
...? - 是否继承所有其他函数(除了constructors / operator =)?
- What is the reason of that ?
- Is there any workaround to inherit the assignment operator ?
- Is it also the case for
operator+=
,operator-=
, ... ? - Are all other functions (apart from constructors/operator=) inherited ?
,我遇到这个问题,因为我在做一些CRTP:
In fact, I encountered this problem as I was doing some CRTP :
template<class Crtp> class Base
{
inline Crtp& operator=(const Base<Crtp>& rhs) {/*SOMETHING*/; return static_cast<Crtp&>(*this);}
};
class Derived1 : public Base<Derived1>
{
};
class Derived2 : public Base<Derived2>
{
};
是否有任何解决方案可以使这项工作?
Is there any solution to get that working ?
编辑:好的,我已经孤立的问题。为什么以下是不工作?如何解决问题?
EDIT : OK, I have isolated the problem. Why the following isn't working ? How to solve the problem ?
#include <iostream>
#include <type_traits>
// Base class
template<template<typename, unsigned int> class CRTP, typename T, unsigned int N> class Base
{
// Cast to base
public:
inline Base<CRTP, T, N>& operator()()
{
return *this;
}
// Operator =
public:
template<typename T0, class = typename std::enable_if<std::is_convertible<T0, T>::value>::type>
inline CRTP<T, N>& operator=(const T0& rhs)
{
for (unsigned int i = 0; i < N; ++i) {
_data[i] = rhs;
}
return static_cast<CRTP<T, N>&>(*this);
}
// Data members
protected:
T _data[N];
};
// Derived class
template<typename T, unsigned int N> class Derived : public Base<Derived, T, N>
{
};
// Main
int main()
{
Derived<double, 3> x;
x() = 3; // <- This is OK
x = 3; // <- error: no match for 'operator=' in ' x=3 '
return 0;
}
推荐答案
赋值运算符在技术上是继承的;但是,它总是被派生类的显式或隐式定义的赋值运算符隐藏(见下面的注释)。
The assignment operator is technically inherited; however, it is always hidden by an explicitly or implicitly defined assignment operator for the derived class (see comments below).
赋值)赋值运算符应该由一个
非静态成员函数来实现,它只有一个参数。因为如果用户未声明
,则为aa类隐式声明复制
赋值运算符operator =
,基类赋值运算符总是
被派生类的副本赋值操作符隐藏。
(13.5.3 Assignment) An assignment operator shall be implemented by a non-static member function with exactly one parameter. Because a copy assignment operator
operator=
is implicitly declared for a a class if not declared by the user, a base class assignment operator is always hidden by the copy assignment operator of the derived class.
你可以实现一个虚拟赋值操作符,基类 operator =
,如下所示:
You can implement a dummy assignment operator which simply forwards the call to the base class operator=
, like this:
// Derived class
template<typename T, unsigned int N> class Derived : public Base<Derived, T, N>
{
public:
template<typename T0, class = typename std::enable_if<std::is_convertible<T0, T>::value>::type>
inline Derived& operator=(const T0& rhs)
{
return Base<Derived, T, N>::operator=(rhs);
}
};
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