One template question

Code:

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template <typename T, typename CONT /*= std::vector*/ >
struct Stack{
    CONT elem;
};

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The above code compiles fine as it should be If I just uncomment the std::vector one it fires compilation errors saying

Code:

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error: expected type-specifier
error: expected '>'

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But I don't find any syntactic or symantic error in the above code. can anybody please point it out.

std::vector of what?

I am guessing that you actually want to write:

Code:

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template <typename T, typename CONT = std::vector<T> >
struct Stack{
    CONT elem;
};

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std::vector is unspecialized.
std::vector<T> in specialized, thus

template <typename T, typename CONT = std::vector>
...does not work.

template < typename T, typename CONT = std::vector<T> >
...works.

and
template <typename T, template<typename> CONT = std::vector>
...also works.

Thanks all understood
I forgot that std::vector does not name a type rather std::vector<SOMETYPE> is a Typename (Silly mistake)

Quote:

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Originally Posted by Elysia

template <typename T, template<typename> CONT = std::vector>
...also works.

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No, it doesn't. std::vector takes an unspecified number of template arguments, but at least 2.

Quote:

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Originally Posted by CornedBee

No, it doesn't. std::vector takes an unspecified number of template arguments, but at least 2.

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What do you mean it takes an unspecified number of template arguments? How is that possible? Can you use the ... operator in template arguments?
As far as I can see, it just takes 2 arguments, one of which is defaulted.

The standard says that the implementation may give the containers any number of template arguments after those specified by the standard, as long as they have defaults.

This, combined with disallowing the substitution of default arguments in template template parameters, has led to many unhappy users.

Tsch. Burned.
Oh well.
At least the second works.