CODEWITHSUNDEEP
c++templates
thundium
thundium

Reputation: 1063

No difference if "typename" is used in template specialization

I have a simple template like

template <typename a, typename b, typename c> 
class myclass
{

};

And I realize that I can specialize the template in both following ways:

template <typename a, typename b>
class myclass<a, b, int> 
{

};

template <typename a, typename b>
class myclass<a, b, typename int> 
{

};

Why there is no difference between "typename int" and "int" ?

Upvotes: 0

Views: 101

Answers (1)

Guillaume Racicot
Guillaume Racicot

Reputation: 41770

There is, it's must be a visual studio bug. Because visual studio don't implement two phase name lookup and has no means of what is a dependent type or not, it allows you to compile it without errors. However, when compiled with gcc, you'll get the following error:

main.cpp:11:33: error: template argument 3 is invalid
class myclass<a, b, typename int>
                    ^

But when using it with dependent names, typename keyword must appear to tell the compiler it's not a value or other thing than a type. In fact, the typename keyword must only appear when it's really needed.

However, there are context where the compiler already know that the expression must yield to a type, like inheritance. If you try to put typename in front of a dependent name there, you'll get an error. Take a look a this code:

template <typename A>
struct MyType : typename std::decay<A>::type {};

This result in that error:

main.cpp:6:17: error: keyword 'typename' not allowed in this context (the base class is implicitly a type)
 struct MyType : typename std::decay<A>::type {};
                 ^

The compiler only need those keywords when there's is an ambiguous statement in a template. Otherwise, you can't put these keywords.

Upvotes: 1

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