当我尝试在模板化 function 中使用迭代器时，“期望在 'typename' 之后有一个限定名称”

Question

I want to make a function that takes a container T (can be vector, map, list...) as template and a T and a Int as arguments, in this function, we're assuming that T is a container of int, and I want to return the first occurence of the int in the container. Here's the function:

template <class T> int & easyfind(T container, int n)
{
    typename T<int>::iterator it;

    for (it = container.begin(); it != container.end(); it++)
        if (*it == n)
            return (*it);
    throw (NotFoundException());
}

But the compiler says "expected a qualified name after 'typename'", and when I replace the typename by class the compiler says "explicit specialization of non-template class 'T'", how can I get this to work?

Answer 1

T is a type, not a template. You need

typename T::iterator it;

to access its iterator type member.

The reason you see code like

std::vector<int>::iterator

is because std::vector is the name of a template, and you need to specify the template parameter. In your case T is already an instantiation of a template, so there is no need to specify the parameter.