在有效和无效类型之间选择 [英] Selecting between valid and non-valid types

问题描述

如果我们有一个模板元功能像 std :: conditional ，我们可以基于布尔编译时条件选择类型。例如：

If we have a template metafunction like std::conditional, we can "select" types based on a boolean compile-time condition. For example:

template < bool CONDITION, typename T, typename U >
struct conditional;

template < typename T, typename U > 
struct conditional < true, T, U >
{
    using type = T;
};

template < typename T, typename U > 
struct conditional < false, T, U >
{
    using type = U;
};

const bool whats_big = sizeof( int ) > sizeof( double );

using bigger_type = typename conditional<whats_big , int , double>::type;

我的问题是：有无法在有效类型和无效类型之间选择？

当前正在实现一个事件类。事件具有发送者参数和事件args的变量号：

Im currently implementing an event class. Events have a sender parameter and variable number of event args:

template<typename SENDER , typename... ARGS>
class event;

因此类型为 void的函数（SENDER&，ARGS& ...） 可以用作事件处理程序。
在这种情况下，处理程序被称为传递一个引用到引发事件的对象（Throught发送方参数）。

另一方面，我想要一种允许发送方成员函数成为事件的处理程序，换句话说，类型 void（SENDER :: *）（ARGS& ...）的函数。

So functions with type void(SENDER& , ARGS&...) could be used as event handlers. In that case the handlers are called passing a reference to the object wich have raised the event (Throught the sender param).
On the other hand, I want a way to allow sender member functions to be handlers of events, in other words, functions of type void(SENDER::*)(ARGS&...).

问题是我不能使用句子：

The problem is that I cannot use a sentence like:

 using handler_type = typename conditional<std::is_class<SENDER>::value,void(SENDER::*)(ARGS&...) , void(SENDER& , ARGS&...)>::type;

因为在 SENDER 类型类型，第一种类型是无效的（使用指向非类型成员的指针）。

Because in the case that SENDER is not a class type, the first type is invalid (Uses a pointer to a member of a non-class type).

推荐答案

可以用额外的间接方法来实现：

You can do it with an extra level of indirection:

template <bool, typename T, typename ...Args>
struct sender_chooser
{
    using type = void(*)(T &, Args &...);
};

template <typename T, typename ...Args>
struct sender_chooser<true, T, Args...>
{
    using type = void (T::*)(Args &...);
};

template <typename T, typename ...Args>
struct sender_type
{
    using type =
        typename sender_chooser<std::is_class<T>::value, T, Args...>::type;
};

用法：

sender_type<MySender, Arg1, Arg2, Arg3>::type

void（MySender :: *）（Arg1& Arg2& Arg3&）如果 MySender 类类型或 void（*）（Sender& amp; Arg1& Arg2& Arg3&）。

（您可能还需要允许联合。）

(You might also want to allow unions.)

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