Re: [Boost-users] Odd Warning in transform_iterator.hpp ?

Subject: Re: [Boost-users] Odd Warning in transform_iterator.hpp ?
From: John M. Dlugosz (mpbecey7gu_at_[hidden])
Date: 2011-07-28 05:48:43

On 7/28/2011 4:10 AM, Nathan Ridge wrote:
>
>> I'm using the current Boost library on Microsoft Visual Studio 2005. I'm getting a
>> warning on line 121 of transform_iterator.hpp, complaining "returning address of local
>> variable or temporary". The line in question is the body of this function:
>>
>> typename super_t::reference dereference() const
>> { return m_f(*this->base()); }
>>
>> My m_f supplied to the template is ordinary enough; it returns a value.
>
> Is your m_f a function object that uses result_of protocol to declare
> its return value?
>
> It may be that while its operator() returns a value, its result<>  is telling
> transform_iterator that it returns a reference.
>
> For example, if its result<>  looks like this:
>
> template<typename>  struct result;
> template<typename F, typename T>
> struct result<F(T)>
> {
>      typedef T type;
> };
>
> then transform iterator will deduce its return type to be a reference:
>
> typedef typename ia_dflt_help<
>              Reference
>            , result_of<UnaryFunc(typename std::iterator_traits<Iterator>::reference)>
>          >::type reference;
>
> and it will use this as the return type of operator*. Hence the warning.
>
> Instead, you need to declare result<>  like this:
>
> template<typename>  struct result;
> template<typename F, typename T>
> struct result<F(const T&)>
> {
>      typedef T type;
> };
>
> (and possibly add a non-const reference version if necessary).
>
> Regards,
> Nate. 		 	   		

Many thanks for the explanation.
Perhaps that should be added to 
<http://www.boost.org/doc/libs/1_47_0/libs/utility/utility.htm#result_of>, or replace the 
struct result<F(T)> in the instructions!