[std-interval] passing by value vs reference

[Guillaume Melquiond]

Le mercredi 05 avril 2006 à 15:37 +0200, Gabriel Dos Reis a écrit :

> | > I would prefer 3 to 4.  I think
> | > the complexity of implementing 4 is not worth it.
> | 
> | I don't see any additional complexity in 4 with respect to 3. As already
> | mentioned by Bill Clarke, you could simply have a typedef member in
> | interval<T> to express the calling convention of the functions of the
> | interval library: typedef interval<T> const &param_type;
> 
> If done that way, in the typical cases, the "T" would then appear in
> non-deducible context. 

Right. This occurred to me when reading Sylvain's reply. However,
although typical, it seems to me this only applies to cases outside the
interval library. Indeed, the functions of the library do not have to
use param_type, since they already know their own calling convention,
and hence T is deducible there. So the issue only applies to
user-defined template functions that would try to benefit from using the
same calling convention.

Anyway, do you know of a way to avoid this issue?
(Please do not suggest macros :-)

Best regards,

Guillaume