Using a templated parameter's value_type - c++

Question

How is one supposed to use a std container's value_type?
I tried to use it like so:
#include <vector>
using namespace std;
template <typename T>
class TSContainer {
private:
T container;
public:
void push(T::value_type& item)
{
container.push_back(item);
}
T::value_type pop()
{
T::value_type item = container.pop_front();
return item;
}
};
int main()
{
int i = 1;
TSContainer<vector<int> > tsc;
tsc.push(i);
int v = tsc.pop();
}
But this results in:
prog.cpp:10: error: ‘T::value_type’ is not a type
prog.cpp:14: error: type ‘T’ is not derived from type ‘TSContainer<T>’
prog.cpp:14: error: expected ‘;’ before ‘pop’
prog.cpp:19: error: expected `;' before ‘}’ token
prog.cpp: In function ‘int main()’:
prog.cpp:25: error: ‘class TSContainer<std::vector<int, std::allocator<int> > >’ has no member named ‘pop’
prog.cpp:25: warning: unused variable ‘v’
I thought this was what ::value_type was for?

Answer 1

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.

Answer 2

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 3

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 4

A fairly common practice is to provide an alias representing the underlying value type for convenience.
template <typename T>
class TSContainer {
private:
T container;
public:
using value_type = typename T::value_type;
void push(value_type& item)
{
container.push_back(item);
}
value_type pop()
{
value_type item = container.pop_front();
return item;
}
};