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问题描述
我打算移植一些代码,这些代码对C ++ 11经常使用boost :: unwrap_reference。该代码调用了很多成员函数,例如
template<类型名T,类型名Y> void initialize(T _t,Y y) { typename boost :: unwrap_reference< T> :: type& t = _t; t.doSomethingNastyWithY(y); } //函数的调用方式如下: struct DoSomething { template< typename Y> void doSomethingNastyWithY(Y y) { //做东西} }; 结构对象{}; DoSomething s; 对象obj; int main() { initialize(s,obj); //取得DoSomething的副本 initialize(boost :: ref(s),obj); //使用DoSomething作为参考}我找不到等同于boost的值: :STL中的unwrap_reference,还有其他直接方法可以做到这一点吗?
编辑:我对示例进行了澄清
解决方案类似于以下内容的东西:
template<类型名称T> struct UnwrapReference; 模板<类型名称T> struct UnwrapReference {typedef T type; } 模板< > 结构体UnwrapReference< std :: reference_wrapper< & > {typedef T type; }虽然未经测试,但这就是您大概能够做到的要点。 / p>
I am tring to port some code, that uses boost::unwrap_reference a lot to C++11. The code calls a lot member functions, like
template< typename T, typename Y> void initialize( T _t, Y y) { typename boost::unwrap_reference< T >::type & t = _t; t.doSomethingNastyWithY( y ); } // The function is called like this struct DoSomething { template<typename Y> void doSomethingNastyWithY(Y y) { // do stuff } }; struct Object {}; DoSomething s; Object obj; int main() { initialize( s, obj ); // Take a copy of DoSomething initialize( boost::ref(s), obj ); // Uses DoSomething as reference }I couldn't find an equivalent to boost::unwrap_reference in the STL, is there an other straight forward way to do this?
EDIT: I clarified the example a bit
解决方案Something along the lines of:
template< typename T > struct UnwrapReference; template< typename T > struct UnwrapReference { typedef T type; } template< > struct UnwrapReference< std::reference_wrapper< T > > { typedef T type; }Untested though, but that's the gist of how you'd probably be able to do it.
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