问题描述
限时送ChatGPT账号..在以下代码中,编译器要求基础类 X默认可构造.但是,如果我从 类 Node 的继承中删除 virtual 关键字,对成员 m_x 的访问当然变得不明确,但不再需要类 X 的默认构造函数.
In the following code, the compiler is requesting the base class X to be default constructible. However, if I remove the virtual keyword from the inheritance of the class Node, the access to the member m_x becomes, of course, ambiguous, but the default constructor for class X is no longer required.
这是什么原因?
#include <iostream>
struct Apply
{
template< typename T >
struct Node : virtual T // this line contains the virtual inheritance
{
template< typename ...Args>
Node( Args... args )
: T( args... )
{}
};
template < typename ...BaseClasses>
struct Inheritance;
template < typename FirstBaseClass, typename ...OtherBaseClasses>
struct Inheritance< FirstBaseClass, OtherBaseClasses... > : FirstBaseClass
, Inheritance< OtherBaseClasses... >
{
template< typename ...Args>
Inheritance( Args... args )
: FirstBaseClass( args... )
, Inheritance< OtherBaseClasses... >( args... )
{
}
};
};
template < >
struct Apply::Inheritance< >
{
template< typename ...Args>
Inheritance( Args... args ){}
};
struct X
{
X(int i){}
int m_x;
};
struct A : Apply::Node< X >
{
A( int i )
: Apply::Node< X >( i )
, m_a( i )
{
}
int m_a;
};
struct B : Apply::Node< X >
{
B( int i )
: Apply::Node< X >( i )
, m_b( i )
{ }
int m_b;
};
struct C : Apply::Node< X >
{
C( int i )
: Apply::Node< X >( i )
, m_c( i )
{ }
int m_c;
};
struct Example : Apply::Inheritance< A, B, C >
{
Example( int i )
: Apply::Inheritance< A, B, C >( i )
{ }
void print( ) const
{
// this line needs the virtual inheritance
std::cout << m_x << std::endl;
std::cout << m_a << std::endl;
std::cout << m_b << std::endl;
std::cout << m_c << std::endl;
}
};
int main()
{
Example ex( 10 );
ex.print( );
return 0;
}
推荐答案
从@Berry 的回答开始,修复代码的唯一方法是编写对虚拟继承的 X 构造函数的显式调用.
Starting from @Berry answer, the only way to fix the code was to code an explicit call to the virtual inherited X constructor.
然而,在A、B或C类中显式调用X的构造是不够的:它必须被调用基本上每个类都涉及到任何级别的继承!
However, it is not enough to explicit call the construction of X in classes A, B, or C: it must be called basically in every class involved in the inheritance at any level!
棘手的是继承<>可变参数模板类:可变参数扩展的每一步都必须提供对X构造函数的显式调用.
The tricky one was the Inheritance<> variadic template class: every step of the variadic expansion must provide the explicit call to the X constructor.
这是在启用 C++11 标志的 MinGW 4.9.2 上运行的代码:
Here is the code that works on MinGW 4.9.2 with enabled C++11 flag:
#include <iostream>
template< typename T, typename V >
struct Node : virtual V
{
using Virtual = V; // Added this line
template< typename ...Args >
Node( Args... args )
: V( args... )
{ }
};
template < typename ...BaseClasses>
struct Inheritance;
template < typename FirstBaseClass, typename ...OtherBaseClasses>
struct Inheritance< FirstBaseClass, OtherBaseClasses... >
: FirstBaseClass
, Inheritance< OtherBaseClasses... >
{
template< typename ...Args>
Inheritance( Args... args )
: FirstBaseClass::Virtual( args... ) // added this line
, FirstBaseClass( args... )
, Inheritance< OtherBaseClasses... >( args... )
{ }
};
template < >
struct Inheritance< >
{
template< typename ...Args >
Inheritance( Args... args )
{ }
};
struct X
{
X(int i)
: m_x( i )
{ }
int m_x;
};
struct A : Node< A, X >
{
A( int i )
: X( i ) // added this line
, Node< A, X >( i )
, m_a( i )
{ }
int m_a;
};
struct B : Node< B, X >
{
B( int i )
: X ( i ) // added this line
, Node< B, X >( i )
, m_b( i )
{ }
int m_b;
};
struct C : Node< C, X >
{
C( int i )
: X ( i ) // added this line
, Node< C, X >( i )
, m_c( i )
{ }
int m_c;
};
struct Example : Inheritance< A, B, C >
{
Example( int i )
: X ( i ) // added this line
, Inheritance< A, B, C >( i )
{ }
void print( ) const
{
// this line needs the virtual inheritance
std::cout << m_x << std::endl;
std::cout << m_a << std::endl;
std::cout << m_b << std::endl;
std::cout << m_c << std::endl;
}
};
int main()
{
Example ex( 10 );
ex.print( );
return 0;
}
这篇关于虚拟继承是否强制基类默认可构造?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!
更多推荐
[db:关键词]
发布评论