If I need to define a template foo
function with a template-template parameter, I usually do the following:
// Notice that the template parameter of class T is unnamed.
template <template <typename> class T> void f() { std::cout << "Yay!
"; }
Notice that the template parameter of the template-template parameter is unnamed, but we can assign a name to this parameter:
// Now the template parameter of class T is named INNER.
template <template <typename INNER> class T> void f(const INNER &inner)
{ std::cout << inner << " Yay!
"; }
This doesn't seems to be usefull at all, because I cannot reffer the INNER
parameter in the function, the code above produces the following error:
error: 'INNER' does not name a type
It surprises me that the typename INNER
doesn't name a type, after all the typename
keyword is there in order to naming a type. Anyway, this is easy to fix though:
// Now INNER is the name of the template parameter of class T and also
// the name of the second template parameter of foo.
template <template <typename INNER> class T, typename INNER> void f(const INNER &inner)
{ std::cout << inner << " Yay!
"; }
// ...
f<std::valarray, int>(666); // Prints "666 Yay!"
But in the end, the INNER
parameter doesn't need a name after all:
// Now the template parameter of class T is unnamed one more time,
// INNER is the name of the second template parameter of foo.
template <template <typename> class T, typename INNER> void f(const INNER &inner)
{ std::cout << inner << " Yay!
"; }
// ...
f<std::valarray, int>(666); // Prints "666 Yay!"
And (sure you already noticed before me) the name in the parameter of the template-template parameter is ignored! It sure have been ignored because if not it should have a name-clash with the second template parameter of foo
, doesn't it?
Another demonstration of the name of the parameter of the template-template parameter being ignored:
// Now T is the name of the template parameter of class T and also
// the name of the template parameter of foo!
template <template <typename T> class T> void f()
{ std::cout << "Yay!
"; }
// ...
f<std::valarray>(); // prints "Yay!"
The type named T
is being used by the template-template parameter and by the template-template itself at the same time? I don't think so, the name into the template-template parameter is ignored AFAIK.
So, what's the question?
- Are my guess correct? The names of named template parameters of template-template parameters are ignored?
- If I'm mistaken and I've misunderstood the whole thing, there's a use for the named parameters into the template-template parameters? Can you provide some useful examples?
As for useful examples on #2 i'm referring to something that only could be achieved using the named template parameters of template-template parameters.
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