Effective C++ item 42: Understand the Two Meanings of typename

program-ape › Uncategorized
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Ever wonder why you need to specify typename when you use nested dependent type name? Consider following example without typename

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template<typename C>
void print2nd(const C& container)
{
    C::const_iterator * x;
    ...
}

This looks like we’re declaring x as a local variable that’s a pointer to a C::const_iterator. But it looks that way only because we “know” that C::const_iterator is a type. But what if C::const_iterator weren’t a type? What if C had a static data member that happened to be named const_iterator, and what if x happened to be the name of a global variable? In that case, the code above wouldn’t declare a local variable, it would be a multiplication of C::const_iterator by x!

This is exactly why typename is needed before nested dependent type - to avoid ambiguity. Now let’s look at two special cases that we don’t need typename, because it won’t have such ambiguity.

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template<typename T>
class Derived: public Base<T>::Nested {   // base class list: typename not
public:                                   // allowed
    explicit Derived(int x)
    : Base<T>::Nested(x)                  // base class identifier in mem
    {                                     // init. list: typename not allowed
        typename Base<T>::Nested temp;    // use of nested dependent type
        ...                               // name not in a base class list or
    }                                     // as a base class identifier in a
    ...                                   // mem. init. list: typename required
};

A practical way of specifying long nested type name is by typedef it

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template<typename IterT>
void workWithIterator(IterT iter)
{
    typedef typename std::iterator_traits<IterT>::value_type value_type;
    value_type temp(*iter);
    ...
}

Last point of typename: when declaring template parameters, class and typename are interchangeable.

Reference:
“Effective C++” Third Edition by Scott Meyers.