限制模板功能，只允许某些类型

Question

在这里说，我有一个简单的模板功能，原则上可以接受所有种类型：

template <class Type> 
std::ostream& operator<< (std::ostream& stream, const Type subject) { 
stream << "whatever, derived from subject\n"; 
return stream; } 

我只是想用这个模板来清点几类，说std :: vector和boost :: array对象。然而，无论何时我将cout用于其他类型甚至基本类型，例如std :: cout < < int（5）;,将是一个编译错误，因为现在有两种可能的运算符< <（std :: ostream，int）的实现，一个是标准C++，另一个是我的模板指定的功能。

我想问一下，是否可以限制我的模板函数，以便它只接受我指定的几种类型？这是如何告诉编译器忽略我的模板当我使用cout < < int（5）。提前致谢。

更清楚，这就是我想做的事：

template <class Type> 
std::ostream& operator<< (std::ostream& stream, const Type subject) { 
if (Type == TypeA or TypeB or TypeC) //use this template and do these {...}; 
else //ignore this template, and use operator<< provided in standard c++ library. 
} 

Answer 1

您可以限制超载是这样的：

template <class T> 
std::ostream& my_private_ostream(std::ostream& stream, const T& data) 
    { <your implementation> } 

template <class T, class A> 
std::ostream& operator<< (std::ostream& stream, const std::vector<T,A>& data) 
    { return my_private_ostream(stream,data); } 

同为std::array S（你应该标记您的问题C++ 11）：

template <class T, size_t N> 
std::ostream& operator<< (std::ostream& stream, const std::array<T,N>& data) 
    { return my_private_ostream(stream,data); } 

---

或者，对于看起来更像您的编辑的解决方案，您可以使用C++ 11 enable_if，尽管我对他们有个人厌恶，因为他们倾向于使代码难以阅读和维护。所以我强烈推荐以前的解决方案。

// Vector type predicate 
template <class T> 
struct is_vector: std::false_type {}; 

template <class T, class A> 
struct is_vector< std::vector<T,A> >: std::true_type {}; 

// Array type predicate 
template <class T> 
struct is_array: std::false_type {}; 

template <class T, size_t N> 
struct is_array< std::array<T,N> >: std::true_type {}; 

// The overload with the syntax you want 
template <class Indexable> 
typename std::enable_if< 
    is_vector<Indexable>::value || is_array<Indexable>::value, 
    std::ostream& 
>::type 
operator<< (std::ostream& stream, const Indexable& data) 
    { <your implementation> } 

Answer 2

使用SFINAE来做你在问什么。

template<typename...> 
struct is_vector: std::false_type{}; 

template<typename T, typename Alloc> 
struct is_vector<std::vector<T, Alloc>>: std::true_type{}; 

template<typename...> 
struct is_array: std::false_type{}; 

template<typename T, std::size_t Size> 
struct is_array<std::array<T, Size>>: std::true_type{}; 

template<typename T> 
struct is_my_ostream_type{ 
    enum { 
     value = is_vector<T>::value || is_array<T>::value 
    }; 
}; 

template< 
     typename T, 
     typename = typename std::enable_if<is_my_ostream_type<T>::value>::type 
> 
std::ostream &operator <<(std::ostream &lhs, const T &rhs){ 
    lhs << "is my special ostream overload"; 
    return lhs; 
} 

但是你可能最终只会为每种类型写一个重载，而不是这样做。

Answer 3

为此编写一个真正通用的解决方案很困难。检查任意类型T与std::vector或std::array的问题是后者不是类，它们是类模板。更糟糕的是，std::array是一个带有非模板参数的类模板，所以你甚至不能拥有一个参数包，它可以容纳std::vector和std::array。

你可以通过在类型中显式包装非类型参数来解决这个问题，但它会变得很难看，速度很快。

这是我提出的一个解决方案，它将默认支持任何没有非模板参数的类或模板类。通过添加包装类型来将非类型参数映射到类型参数，可以支持具有非类型模板参数的模板类。

namespace detail{ 
    //checks if two types are instantiations of the same class template 
    template<typename T, typename U> struct same_template_as: std::false_type {}; 
    template<template<typename...> class X, typename... Y, typename... Z> 
    struct same_template_as<X<Y...>, X<Z...>> : std::true_type {}; 

    //this will be used to wrap template classes with non-type args 
    template <typename T> 
    struct wrapImpl { using type = T; }; 

    //a wrapper for std::array 
    template <typename T, typename N> struct ArrayWrapper; 
    template <typename T, std::size_t N> 
    struct ArrayWrapper<T, std::integral_constant<std::size_t, N>> { 
     using type = std::array<T,N>; 
    }; 

    //maps std::array to the ArrayWrapper 
    template <typename T, std::size_t N> 
    struct wrapImpl<std::array<T,N>> { 
     using type = ArrayWrapper<T,std::integral_constant<std::size_t,N>>; 
    }; 

    template <typename T> 
    using wrap = typename wrapImpl<typename std::decay<T>::type>::type; 

    //checks if a type is the same is one of the types in TList, 
    //or is an instantiation of the same template as a type in TempTList 
    //default case for when this is false 
    template <typename T, typename TList, typename TempTList> 
    struct one_of { 
     using type = std::false_type; 
    }; 

    //still types in the first list to check, but the first one doesn't match 
    template <typename T, typename First, typename... Ts, typename TempTList> 
    struct one_of<T, std::tuple<First, Ts...>, TempTList> { 
     using type = typename one_of<T, std::tuple<Ts...>, TempTList>::type; 
    }; 

    //type matches one in first list, return true 
    template <typename T, typename... Ts, typename TempTList> 
    struct one_of<T, std::tuple<T, Ts...>, TempTList> { 
     using type = std::true_type; 
    }; 

    //first list finished, check second list 
    template <typename T, typename FirstTemp, typename... TempTs> 
    struct one_of<T, std::tuple<>, std::tuple<FirstTemp, TempTs...>> { 
     //check if T is an instantiation of the same template as first in the list 
     using type = 
      typename std::conditional<same_template_as<wrap<FirstTemp>, T>::value, 
       std::true_type, 
       typename one_of<T, std::tuple<>, std::tuple<TempTs...>>::type>::type; 
    }; 
} 

//top level usage 
template <typename T, typename... Ts> 
using one_of = typename detail::one_of<detail::wrap<T>,Ts...>::type; 

struct Foo{}; 
struct Bar{}; 

template <class Type> 
auto operator<< (std::ostream& stream, const Type subject) 
    //is Type one of Foo or Bar, or an instantiation of std::vector or std::array 
    -> typename std::enable_if< 
       one_of<Type, std::tuple<Foo,Bar>, std::tuple<std::vector<int>,std::array<int,0>> 
     >::value, std::ostream&>::type 
{ 
    stream << "whatever, derived from subject\n"; 
    return stream; 
} 

请不要使用这个，这太可怕了。的[C++模板，仅接受某些类型]（

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