David Abrahams <dave <at> boost-consulting.com> writes:
on Fri Jul 27 2007, "JOAQUIN LOPEZ MU?Z" <joaquin-AT-tid.es> wrote:
According to the MPL reference, a placeholder expression X<a1,..,an> must satisfy (among others) the following condition:
"All of X's template parameters, including the default ones, are types."
According to that the following should then compile without errors: [...] I don't see how you can conclude the above will compile without error just from (a part of) the definition of placeholder expression.
Are you saying that since foo<_1> is not a placeholder expression, when applied to int it should be treated as a metafunction class?
Yes, this is exactly my point. Another way in which foo<_1> can be forced to not be treated as a ph expression is this: // no of parms > BOOST_MPL_LIMIT_METAFUNCTION_ARITY template<typename T, typename=int,typename=int,typename=int, typename=int,typename=int> struct foo { template<typename Q> struct apply{ typedef Q type; }; }; You can try yourself this variation and see that the program really resolves to t1 <- int. But for some reason, the int=0 variation does not do the same, at least in GCC 3.4.4. Joaquín M López Muñoz Telefónica, Investigación y Desarrollo PS: I'd greatly appreciate if you could you try the snippet (the int=0 version) with other compilers --my current compiler count is so very limited. Thx! PPS: In case you're wondering, this is not a purely academic question: I've got a real world scenario where I want to prevent foo<x> from being treated as a ph expression, namely when foo<x> is a metafunction class parameterized by a lambda expression (i.e. a ph expression or metafunction class) x.