Why is the keyword "typename" needed before qualified dependent names, and not before qualified independent names?

Solution 1:

A name in C++ can pertain to three different tiers of entities: Types, values, and templates.

struct Foo
{
    typedef int A;                   // type
    static double B;                 // value
    template <typename T> struct C;  // template
};

The three names Foo::A, Foo::B and Foo::C are examples of all three different tiers.

In the above example, Foo is a complete type, and so the compiler knows already what Foo::A etc. refer to. But now imagine this:

template <typename T> struct Bar
{
    T::A x;
};

Now we are in trouble: what is T::A? if T = Foo, then T::A = int, which is a type, and all is well. But when T = struct { static char A; };, then T::A is a value, which doesn't make sense.

Therefore, the compiler demands that you tell it what T::A and T::B and T::C are supposed to be. If you say nothing, it is assumed to be a value. If you say typename, it is a typename, and if you say template, it is a template:

template <typename T> struct Bar
{
    typename T::A x;    // ah, good, decreed typename

    void foo()
    {
        int a = T::B;   // assumed value, OK

        T::template C<int> z;  // decreed template
        z.gobble(a * x);
    }
};

Secondary checks such as whether T::B is convertible to int, whether a and x can be multiplied, and whether C<int> really has a member function gobble are all postponed until you actually instantiate the template. But the specification whether a name denotes a value, a type or a template is fundamental to the syntactic correctness of the code and must be provided right there during the template definition.

Solution 2:

Because in the nontemplate foo, you are doing a valid multiplication operation (assuming you declared iter before that use). Try to omit the star, and you will get a compiler error. The int hides the class.

The typename keyword does not prevent that hiding. Only gcc implements it incorrectly to do so. So if you try to instantiate your function template with A as the type, then you will get a compile error, because the name specified afteer typenake will refer to a nontype on any standard conformant compiler.