是否可以在运行时迭代 mpl::vector 而无需实例化向量中的类型?

2021-12-26 00:00:00 foreach vector c++ boost boost-mpl

通常，我会使用 boost::mpl::for_each<>() 来遍历 boost::mpl::vector，但这需要一个函子模板函数声明如下:

Generally, I would use boost::mpl::for_each<>() to traverse a boost::mpl::vector, but this requires a functor with a template function declared like the following:

templatevoid operator()(T&){T::staticCall();}

我的问题是我不希望对象 T 被 for_each<> 实例化.我根本不需要 operator() 中的 T 参数.有没有办法实现这一点，或者 for_each<> 的替代方法，它不会将 T 类型的对象传递给模板函数?

My problem with this is that I don't want the object T to be instantiated by for_each<>. I don't need the T parameter in the operator() at all. Is there a way to accomplish this, or an alternative to for_each<> that doesn't pass an object of type T to the template function?

最理想的是，我希望 operator() 定义如下所示:

Optimally, I would like the operator() definition to look like this:

templatevoid operator()(){T::staticCall();}

当然，我不希望在调用之前对 T 进行实例化.也欢迎任何其他提示/建议.

And of course, I don't want T to be instantiated at all prior to the call. Any other tips/suggestions are also welcome.

推荐答案

有趣的问题！据我所知，Boost.MPL 似乎没有提供这样的算法.但是，使用迭代器编写自己的代码应该不会太困难.

Interesting question! As far as I can tell, Boost.MPL does not seem to provide such an algorithm. However, writing your own should not be too difficult using iterators.

这是一个可能的解决方案:

Here is a possible solution:

#include <boost/mpl/begin_end.hpp>
#include <boost/mpl/next_prior.hpp>
#include <boost/mpl/vector.hpp>

using namespace boost::mpl;


namespace detail {

template < typename Begin, typename End, typename F >
struct static_for_each
{
    static void call( )
    {
        typedef typename Begin::type currentType;

        F::template call< currentType >();
        static_for_each< typename next< Begin >::type, End, F >::call();
    }
};


template < typename End, typename F >
struct static_for_each< End, End, F >
{
    static void call( )
    {
    }
};

} // namespace detail


template < typename Sequence, typename F >
void static_for_each( )
{
    typedef typename begin< Sequence >::type begin;
    typedef typename end< Sequence >::type   end;

    detail::static_for_each< begin, end, F >::call();
}

[命名可能不是很好，但很好...]

[The naming may not be very well chosen, but well...]

以下是您将如何使用此算法:

Here is how you would use this algorithm:

struct Foo
{
    static void staticMemberFunction( )
    {
        std::cout << "Foo";
    }
};


struct Bar
{
    static void staticMemberFunction( )
    {
        std::cout << "Bar";
    }
};


struct CallStaticMemberFunction
{
    template < typename T >
    static void call()
    {
        T::staticMemberFunction();
    }
};


int main()
{
    typedef vector< Foo, Bar > sequence;

    static_for_each< sequence, CallStaticMemberFunction >(); // prints "FooBar"
}

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