仅为特定的模板化类启用模板

med*_*106 6 c++ static-assert enable-if variadic-templates c++11

这个问题的灵感来自我之前的问题没有模板参数扣除参数包.

请考虑以下代码示例:

#include <memory>
#include <string>

template<typename... FArgs>
class Callback
{
    public:
    class Handle{};
};

class BaseCallbackHandle
{
};

using TypeErasedCallbackHandle = std::unique_ptr<BaseCallbackHandle>;

template<typename H>
TypeErasedCallbackHandle makeTypeErasedCallbackHandle( H handle)
{
   return {};
}

int main()
{
    Callback<int>::Handle h;
    std::string s;
    makeTypeErasedCallbackHandle(h); //should compile fine
    makeTypeErasedCallbackHandle(s); //should raise a compile error
}
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另见http://coliru.stacked-crooked.com/a/5f2a2e816eef6afd

函数模板makeTypeErasedCallbackHandle现在将任何类作为输入参数.有没有办法确保(例如使用static-assert或enable_if),只允许Callback<FArgs...>::Handle(有任何FArgs)HCallback<int>::Handle应该编译的例子,但是std::string会失败.

And*_*rew 1

在类中定义一个类型Handle,并在内部引用该类型makeTypeErasedCallbackHandle()

#include <memory>
#include <string>

template <typename... FArgs>
struct Callback {
    struct Handle {
        using callback_type = Callback<FArgs...>;    
    };
};

struct BaseCallbackHandle {
};

using TypeErasedCallbackHandle = std::unique_ptr<BaseCallbackHandle>;

template <typename H>
TypeErasedCallbackHandle makeTypeErasedCallbackHandle(H handle) {
    using callback_type = typename H::callback_type;

    return {};
}

int main() {
    Callback<int>::Handle h;
    std::string s;
    makeTypeErasedCallbackHandle(h); //should compile fine
    makeTypeErasedCallbackHandle(s); //should raise a compile error
}
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实例

对于任何H未定义嵌套类型的实例,这将在实例化期间失败。


只需多一点努力,您就可以static_assert向客户端生成有意义的消息,同时通过类型特征提高解决方案的灵活性。这样做的优点是callback_impl::is_callback可以专门用于任意句柄类型:

#include <memory>
#include <string>

namespace callback_impl {

struct callback_identification_type {};

template <typename T, typename = void>
struct is_callback : std::false_type {};

template <typename T>
struct is_callback<T,
    std::enable_if_t<std::is_same<typename T::callback_id_type,
                     callback_identification_type>::value>>
 : std::true_type {};

}

template <typename... FArgs>
struct Callback {
    struct Handle {
        using callback_id_type = callback_impl::callback_identification_type;    
    };
};

struct BaseCallbackHandle {
};

using TypeErasedCallbackHandle = std::unique_ptr<BaseCallbackHandle>;

template <typename H>
TypeErasedCallbackHandle makeTypeErasedCallbackHandle(H handle) {
    static_assert(callback_impl::is_callback<H>::value,
                  "The handle type is not a member of a recognised Callback<T...>");
    return {};
}

int main() {
    Callback<int>::Handle h;
    std::string s;
    makeTypeErasedCallbackHandle(h); //should compile fine
    makeTypeErasedCallbackHandle(s); //should raise a compile error

    return 0;
}
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实例

输出:

g++ -std=c++14 -O2 -Wall -Wno-unused-local-typedefs -pedantic -pthread main.cpp && ./a.out
main.cpp: In instantiation of 'TypeErasedCallbackHandle makeTypeErasedCallbackHandle(H) [with H = std::__cxx11::basic_string<char>; TypeErasedCallbackHandle = std::unique_ptr<BaseCallbackHandle>]':
main.cpp:41:35:   required from here
main.cpp:32:5: error: static assertion failed: The handle type is not a member of a recognised Callback<T...>
     static_assert(callback_impl::is_callback<H>::value,
     ^~~~~~~~~~~~~
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