基于范围的循环C++ 11用于范围(L,R)

Question

对于远程整数序列,C++ 11没有基于范围的循环.

for(auto e : {0..10} )  // wouldn't compile!!!

所以我决定模拟它.

template< class T , bool enable = std::is_integral<T>::value >
struct range_impl
{
    struct iterator
    {
        constexpr T operator * ()const noexcept { return value; }
        iterator& operator ++()noexcept { ++value; return *this; }

        friend
        constexpr bool operator != (const iterator & lhs, const iterator rhs ) noexcept
        {
            return lhs.value != rhs.value;
        }
        T value;
    };

    constexpr iterator begin()const noexcept { return { first }; }
    constexpr iterator end  ()const noexcept { return { last  }; } 

    T first;
    T last ;
}; 




template< class T >
range_impl<T>  range(T first , T last) noexcept
{
    return {first, last};
}

int main(){
     // print numbers in [ 0..10 ), i.e. 0 1 2 3 4 5 6 7 8 9
     for(auto e : range(0,10) ) std::cout << e <<  ' ';
     std::cout << std::endl;
}

问:如何为ForwardIterators推广此方法？

例:

template< class ForwardIterator, class T >
bool find(ForwardIterator first, ForwardIterator last, T const& value)
{
     for(auto e: range(first, last) ) if (e == v) return true;
     return false;
}

Answer 1

专业化

template< class Iterator>
struct range_impl<Iterator, false>
{
    range_impl(Iterator first, Iterator last)
    : first(first), last(last)
    {}

    constexpr Iterator begin()const noexcept { return { first }; }
    constexpr Iterator end  ()const noexcept { return { last  }; }

    Iterator first;
    Iterator last ;
};

测试:

int main(){
     for(auto e : range(0,10) ) std::cout << e <<  ' ';
     std::cout << std::endl;
     const char* a[] = { "Say", "hello", "to", "the", "world" };
     for(auto e : range(a, a + 5) ) std::cout << e <<  ' ';
     std::cout << std::endl;
}