这里说我有一个原则上可以接受所有类型的简单模板函数:

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

我只想用这个模板来计算一些类型，比如 std::vector 和 boost::array 对象。但是，每当我将 cout 用于其他类型甚至是基本类型时，例如std::cout < int(5);，将是一个编译错误，因为现在有两种可能的=""><(std::ostream, int)="" 实现，一种是在标准="" c++="">

请问，是否可以限制我的模板功能，使其只接受我指定的几种类型？这就是当我使用 cout < int(5)="">

更清楚地说，这就是我想要做的:

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.
}

最佳答案

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

您可以通过在类型中显式包装非类型参数来稍微解决这个问题，但它变得丑陋、快速。

这是我想出的一个解决方案，它默认支持任何没有非类型模板参数的类或模板类。通过添加包装类型以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; 
}

请不要使用它，它太可怕了。

Live Demo

关于c++ - 限制模板函数，只允许某些类型，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/32267324/