Re: [boost] [lambda] Wrong return type for dereference for pointer to member

From: Jaakko Järvi (jarvi_at_[hidden])
Date: 2006-11-29 23:46:27

On Nov 29, 2006, at 3:43 PM, Rodolfo Lima wrote:

> Lambda's return type deduction for a pointer to member action  
> doesn't honour
> the object constness.
> Regardless of the object constness, the return type is always a non  
> const
> reference to the member type.
>
> The following code illustrates the issue:
>
> struct aux
> {
>     aux() : b(1) {}
>     int b;
> };
>
> struct aux_ptr
> {
>     aux_ptr(aux *ptr=NULL) : m_ptr(ptr) {}
>     ~aux_ptr() { delete m_ptr; }
>
>     template <class R> R &operator->*(R aux::*m) { return m_ptr->*m; }
>     template <class R> const R &operator->*(R aux::*m) const { return
> m_ptr->*m; }
> private:
>     aux *m_ptr;
> };
>
> int main()
> {
>     // This compiles nicely
>     aux_ptr ptr(new aux());
>     cout << (_1->*&aux::b)(ptr) << endl;
>
Hmm... why does this even compile? strange.

>     // This fails to compile.
>     const aux_ptr cptr(new aux());
>     cout << (_1->*&aux::b)(cptr) << endl;
>     return 0;
> }
>
Both versions above are delayed calls to user-defined operator->* and  
the in general the library
won't guess what the result type should be.
In the latter case the lib would have to guess that constness of  
aux_ptr means
that R should be constified in the return, which is too much guess work.

This should work:

     std::cout << ret<const int&>(_1->*&aux::b)(cptr) << std::endl;

Cheers,

   Jaakko

> The second part should compile if other_action<member_pointer_action>
> honoured _1's constness.
>
> Thanks,
> Rodolfo Lima
>
>
>
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