Boost C++ Libraries: Ticket #6278: mem_fn has a different result_type

giecrilj@… — Fri, 16 Dec 2011 11:55:29 GMT

It is unwise for mem_fn to define result_type at all. The type of the member need not even be copyable. Instead, it should be arranged so that

boost:: result_of < F (U &) >:: type be type_of_the_member &
boost:: result_of < F (U const &) >:: type be type_of_the_member const &

Peter Dimov — Fri, 16 Dec 2011 13:52:55 GMT

It's true that result_type is defined as a const reference instead of a plain type, but what does this have to do with whether the type is copyable or not? result_of<F(U&)> is not necessarily M&, by the way. U can be a pointer or a smart pointer to a const pointee. What specific problem are you having?

giecrilj@… — Fri, 16 Dec 2011 18:50:05 GMT

If result_type were plain, as it allegedly is, and the type of the member were noncopyable then returning it from a function would be a problem.

Specific problem:

you cannot copy into a transform_iterator when the transform selects a reference to a member by means of mem_fn.

If I may digress a bit, I suppose that the reason why the standard does not provide an analogue for mem_fun_ref for data members is that accessing data members is discouraged in C++. Having come to that conclusion, I added the necessary accessors to the underlying class and used just mem_fun_ref, eradicating mem_fn completely.

viboes — Mon, 22 Oct 2012 22:01:37 GMT

The error is normal as you have declared twice the type r :(

 typedef typename F:: result_type r;
 typedef type_of_the_member       r;