本文整理汇总了C++中range::back方法的典型用法代码示例。如果您正苦于以下问题:C++ range::back方法的具体用法?C++ range::back怎么用?C++ range::back使用的例子?那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类range
的用法示例。
在下文中一共展示了range::back方法的4个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于系统推荐出更棒的C++代码示例。
示例1: is_mergeable
inline bool is_mergeable (const range<iter1_t> &src1,
const range<iter2_t> &src2, compare comp )
{ //---------------------------- begin ------------------------------------
typedef typename iterator_traits<iter1_t>::value_type type1 ;
typedef typename iterator_traits<iter2_t>::value_type type2 ;
static_assert ( std::is_same<type1, type2>::value,
"Incompatible iterators\n");
//---------------------------- begin --------------------------------------
return comp ( *(src2.front()), *(src1.back()));
};
示例2: join
inline range<entity<Segments ... >> join(range<entity<Segments ... >> const & one, range<entity<Segments ... >> const & two) noexcept
{
return { std::min(one.offset(), two.offset()), (std::max(one.back(), two.back()) + 1) - std::min(one.offset(), two.offset()) };
}
示例3: intersects
inline type::bool_t intersects(range<entity<Segments ... >> const & one, range<entity<Segments ... >> const & two) noexcept
{
return (one.front() <= two.back()) && (two.front() <= one.back());
}
示例4:
inline range<entity<Segments ... >> take_last(range<entity<Segments ... >> const & object, type::index_t const count) noexcept
{
return { object.back() - ((count % (object.length() + 1)) - 1), count % (object.length() + 1) };
}