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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