Template aliases are very convenient in simplifying types like typename F <T>::type
to just F <T>
, where T
and type
are types.
I would like to do the same for templates like F <T>::map
, i.e., simplify them to F <T>
, where T
and map
are template structs or aliases.
For instance, consider the following definitions:
template <bool B>
using expr = std::integral_constant <bool, B>;
template <bool B>
using _not = expr <!B>;
template <template <typename> class F>
struct neg_f
{
template <typename T>
using map = _not <F <T>{}>;
};
template <typename T>
pred = expr < /* ... T ... */ >; // e.g., pred = expr <true>;
template <template <typename> class F>
struct fun;
Now the following works:
fun <neg_f <pred>::map>
This would be much more convenient, but it fails:
template <template <typename> class F>
using neg = neg_f <F>::map;
fun <neg <pred> >
(It also fails with neg = neg_f <F>::template map
, even if map
is defined as a struct). It appears that the definition of neg
above would rather have to be like a "template template alias"
template <template <typename> class F>
template <typename T>
using neg = neg_f <F>::template map <T>;
but apparently there is no such thing.
So, is there any solution or should I stay with neg_f <pred>::map
?
At first consider using
typename
keyword to state that this is a nested type no matter if one is a type (e.g. struct, class, etc), template type, typedef or alias.Alias specification requires you to use a type-id to specify a previously defined type. In this particular case correct using of type-id will look like this:
What you try to do originally is to use template-name
neg_f<F>::map
as a type-id. This is not correct.Probably you want to deduce somehow the
T
parameter fromF
to be used attemplate map<T>
but this is not applicable to your final use-casefun<neg<pred>>
whereT
is not resolved.