On 5/9/06, Tobias Schwinger <tschwinger@neoscientists.org> wrote:
Do you already have possible overloads and return types for convert() in mind?
Sure: 'convert' returns an expression, just like any other operator. That expression will be something like a reference wrapper for the argument.
When the full compound expression is evaluated the convert expression protects its component expression from result type propagation.
Best case, it's all it does -- the evaluation machinery can view the result of the evaluation as a value of the expected domain.
void f( rational const & ) ; void f( integer const & ) ; integer a , b ; f( convert( a ) / b ) ; what is going to happen ? or do you expect end-users to provide overloads like void f( integer_expression_stub const & ) ; ? then they are bound to a particular implementation of integer. am i missing something? *** i am also not convinced the proposed evaluation mechanism is really useful: it may look better than rational r = rdiv( a , b ) ; to some, but looks worse for others, and it goes against existing practice; but maybe i'm just too conservative br, andras