模板化类的模板化构造函数的显式模板专门化 [英] Explicit template specialization for templated constructor of templated class

问题描述

此问题可能与此重叠： C +模板化类的模板化构造函数的显式模板专门化。
然而，我没有在该线程中找到解决方案。

我有一个模板化的类与模板化的构造函数：

 模板< typename首先，typename ... Rest> class var {
 public：
 template< typename T> var（T& t）{
 std :: cout< 一般< std :: endl; 
} 
}; 
  

但是，如果这个类被实例化为一个与参数相同类的对象（即， d喜欢调用copy-（或move-）构造函数）应该做什么具体的。所以我尝试以下：

 模板< typename首先，typename ... Rest>模板<> 
 var< First，Rest ...> :: var（var< First，Rest ...>& v）{
 std :: cout< copy<< std :: endl; 
} 
  

当尝试使用g ++ 4.6编译时得到
错误： '''token
之前无效的显式专门化
错误：封闭类模板没有明确专用
之前的错误混淆了

问题，我不得不明确说明哪个类我想专门化的构造函数...

但是，我希望它变得清楚我想做什么。任何想法如何？

解决方案

模板构造函数不是复制构造函数：

  template< typename首先，typename ... Rest> class var {
 public：
 var（）{}; 
 var（const var& v）{
 std :: cout< copy<< std :: endl; 
} 
 template< typename T> 
 var（const T& t）{
 std :: cout< 一般< std :: endl; 
} 
}; 
 
 int main（）
 {
 var< int> i0; 
 var< int> i1（i0）; 
 var< int> i2（Hello）; 
} 
  

给予

  copy 
 general 
  

注意： / p>

您尝试将非复制构造函数专门用作复制构造函数失败。

12.8：

类X的非模板构造函数是一个拷贝构造函数，如果它的
第一个参数是X& const X& volatile X&或const volatile
X&，没有其他参数，或者所有其他
参数都有默认参数（8.3.6）。

This question might overlap with this one: C++ explicit template specialization of templated constructor of templated class . However, I didn't find a solution in that thread.

I have a templated class with templated constructor:

template<typename First, typename ... Rest> class var {
    public:
        template<typename T> var(T& t) {
            std::cout << "general" << std::endl;
        }
};

But in case this class is instantiated with an object of the same class as parameter (i.e., we'd like to call the copy- (or move-) constructor) something specific should be done. So I tried the following:

template<typename First, typename ... Rest> template<> 
var<First, Rest...>::var(var<First, Rest...>& v) {
    std::cout << "copy" << std::endl;
}

When trying to compile this with g++ 4.6 I get error: invalid explicit specialization before ‘>’ token error: enclosing class templates are not explicitly specialized confused by earlier errors, bailing out

I see the problem, I would have to say explicitly for which class I would like to specialize the constructor...

However, I hope it became clear what I want to do. Any ideas how?

解决方案

A template constructor is no copy constructor:

template<typename First, typename ... Rest> class var {
    public:
    var() {};
    var(const var& v) {
        std::cout << "copy" << std::endl;
    }
    template<typename T>
    var(const T& t) {
        std::cout << "general" << std::endl;
    }
};

int main()
{
    var<int> i0;
    var<int> i1(i0);
    var<int> i2("Hello");
}

Gives

copy
general

Note: Added some const

Your attempt to specialize a non copy constructor as copy constructor is failing.

12.8:

A non-template constructor for class X is a copy constructor if its first parameter is of type X&, const X&, volatile X& or const volatile X&, and either there are no other parameters or else all other parameters have default arguments (8.3.6).

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