Question

我遇到函数boost::make_iterator_range

的别名

（我希望隐藏一个别名后面的提升，以防这个特定的库在将来的某个时候被采用到标准中。）

有什么方法可以使它起作用吗？

#include <boost/range/iterator_range.hpp>

void Foo()
{
}

template< typename T >
void Bar()
{ }

template< typename T >
void Bar(char c)
{ }

void (&FooAlias)() = Foo;           // ok
void (&BarAlias)() = Bar<int>;      // ok

// boost::iterator_range<const size_t*> (&MakeIterRangeAlias)(const size_t*,const size_t*) = 
//  boost::make_iterator_range<const size_t*>;  // not ok

int main(int argc, char** argv)
{
    const size_t v[] = { 3, 5, 1, 5, 29, 15 };

    boost::iterator_range<const size_t*> r 
        = boost::make_iterator_range( std::begin( v ), std::end( v )); // want to alias this

    return 0;
}

错误消息是：

In file included from /usr/include/boost/iterator/iterator_categories.hpp:15:0,
                 from /usr/include/boost/iterator/detail/facade_iterator_category.hpp:7,
                 from /usr/include/boost/iterator/iterator_facade.hpp:14,
                 from /usr/include/boost/range/iterator_range_core.hpp:23,
                 from /usr/include/boost/range/iterator_range.hpp:13,
                 from sandbox.cpp:2:
/usr/include/boost/mpl/eval_if.hpp: In instantiation of ‘boost::mpl::eval_if_c<true, boost::range_const_iterator<const long unsigned int*>, boost::range_mutable_iterator<const long unsigned int* const> >’:
/usr/include/boost/range/iterator.hpp:63:63:   instantiated from ‘boost::range_iterator<const long unsigned int* const>’
sandbox.cpp:20:10:   instantiated from here
/usr/include/boost/mpl/eval_if.hpp:60:31: error: no type named ‘type’ in ‘boost::mpl::eval_if_c<true, boost::range_const_iterator<const long unsigned int*>, boost::range_mutable_iterator<const long unsigned int* const> >::f_ {aka struct boost::range_const_iterator<const long unsigned int*>}’
/usr/include/boost/mpl/eval_if.hpp: In instantiation of ‘boost::mpl::eval_if_c<false, boost::range_const_iterator<const long unsigned int*>, boost::range_mutable_iterator<const long unsigned int*> >’:
/usr/include/boost/range/iterator.hpp:63:63:   instantiated from ‘boost::range_iterator<const long unsigned int*>’
sandbox.cpp:20:10:   instantiated from here
/usr/include/boost/mpl/eval_if.hpp:60:31: error: no type named ‘type’ in ‘boost::mpl::eval_if_c<false, boost::range_const_iterator<const long unsigned int*>, boost::range_mutable_iterator<const long unsigned int*> >::f_ {aka struct boost::range_mutable_iterator<const long unsigned int*>}’
sandbox.cpp:20:10: error: invalid initialization of non-const reference of type ‘void (&)(const size_t*, const size_t*) {aka void (&)(const long unsigned int*, const long unsigned int*)}’ from an rvalue of type ‘<unresolved overloaded function type>’
make: *** [sandbox] Error 1

Answer 1

使用函数指针是一种对函数进行别名的次优方法。它不像原版那样灵活（它不再是模板），你现在需要知道函数的确切签名，这可能是稳定的，也可能是不稳定的。而是尝试这种方法。

template< typename ... Args >
auto MakeIterRangeAlias( Args&& ... args ) -> decltype( /* copy return line here */ )
{
   return boost::make_iterator_range( std::forward<Args>(args)... );
}

由于您的工作几乎没有，别名支持原始的确切签名。即使它发生了巨大的变化，你仍然会被设定。此外，与函数指针方法不同，优化器将能够简单地内联MakeIterRangeAlias，以便不存在运行时开销。

是否可以为boost :: make_iterator_range添加别名？

1 个答案: