g++ "is not a type" error

Question

Writing a templated function, I declared:

template <typename T>
T invertible(T const& container, T::size_type startIndex, T::size_type endIndex);

Compiling with g++ 4.0.1 I got the error:

error: 'T::size_type' is not a type

Answer 1

You need to add typename.

I.e.

template <typename T>
T invertible(T const& container, typename T::size_type startIndex, typename T::size_type endIndex);

Without having any information on your type T, the compiler needs to know that T::size_type designates a type.

From the standard, section 14.6.2:

A name used in a template declaration or definition and that is dependent on a template-parameter is assumed not to name a type unless the applicable name lookup finds a type name or the name is qualified by the keyword typename.

Answer 2

It turned out I needed to specify that T::size_type was a typename. Why is that?

template <typename T>
T invertible(T const& container, typename T::size_type startIndex, typename T::size_type endIndex);

Answer 3

Because during parsing the template declaration, T is not known. So the compiler doesn't know if T::size_type exists at all. It may refer to a static variable, for example. When you later use the template, T is of course known, but the error happens earlier. And please use something less ancient than gcc 4.0.1 ;-)

Edit: If you compile it with -fpermissive, the compiler probably chews your code, but he will give a warning.

Answer 4

As you found out T::size_type needs to be prefixed with typename. Why?

From "C++ Templates: The Complete Guide"

The language definition resolves this problem by specifying that in general a dependent qualified name does not denote a type unless that name is prefixed with the keyword typename.

... The typename prefix to a name is required when the name

1. Appears in a template
2. Is qualified
3. Is not used as a list of base class specifications or in a list of member initializations introducing a constructor definition
4. Is dependent on a template parameter

Furthermore the typename prefix is not allowed unless at least the first three previous conditions hold.