Why does iterator type deduction fail? [duplicate]

Answer 1

The main reason it doesn't work is because the sandard says that the T here is in a non-deduced context. The reason why the context is not deduced is because when you pass some type to the function, the compiler would have to instantiate every single possible std::vector (including for types not present in this particular translation unit) in order to try to find one which had a corresponding type.

Of course, in case of std::vector, the compiler could contain some magic to make this work, since the semantics of the class are defined by the standard. But generally, TemplateClass<T>::NestedType can be a typedef to literally anything, and there's nothing the compiler can do about it.

Answer 2

Simple.

struct X {};
template<> class std::vector<X> {
    typedef std::vector<int>::iterator iterator;
};

Oops.

Templates are Turing-Complete. You are asking the compiler to infer the arguments from the results. This is impossible in the general case, even ignoring the possibility of non-one-to-one correspondence.

Typically, you take the iterator type itself as template parameter. This permits other random-access iterators like deque, circular buffer, etc.