Using a templated parameter's value_type

Question

How is one supposed to use a std container\'s value_type?
I tried to use it like so:

#include 

using namespace std;

template 

        

Answer 1

Use the typename keyword to indicate that it's really a type.

void push(typename T::value_type& item)

typename T::value_type pop()

Answer 2

Here is a full implementation of the accepted answers above, in case it helps anyone.

#include <iostream>
#include <list>

template <typename T>
class C1 {
  private:
    T container;
    typedef typename T::value_type CT;
  public:
    void push(CT& item) {
        container.push_back(item);
    }

    CT pop (void) {
        CT item = container.front();
        container.pop_front();
        return item;
    }
};

int main() {
    int i = 1;
    C1<std::list<int> > c;
    c.push(i);
    std::cout << c.pop() << std::endl;
}

Answer 3

You have to use typename:

typename T::value_type pop()

and so on.

The reason is that the compiler cannot know whether T::value_type is a type of a member variable (nobody hinders you from defining a type struct X { int value_type; }; and pass that to the template). However without that function, the code could not be parsed (because the meaning of constructs changes depending on whether some identifier designates a type or a variable, e.g.T * p may be a multiplication or a pointer declaration). Therefore the rule is that everything which might be either type or variable and is not explicitly marked as type by prefixing it with typename is considered a variable.