Mailinglist Archive: opensuse-programming (8 mails)

[John D Lamb <J.D.Lamb@xxxxxxxxxxxxxx>]

On Sun, 2007-01-21 at 14:30 -0500, Jerry Feldman wrote:
> I have a template class that contains a container:
> template <class T>
> class MyClass
> {	
> 	:
> private:
> 	std::list<T> ilist;
> };
> 
> I have no problem accessing the STL container using subscript,
> but I am unable to declare an iterator, as an example:
> void MyClass<T>::erase(size_t n)
> {
> 	    std::list<T>::iterator it;
> ...

Try replacing this last line with

        typename std::list<T>::iterator it;

The keyword typename has to be used whenever a name that depends on a
template parameter is a type. In this case std::list<T>::iterator is a
type and so we have to qualify it with typename.

I think the rationale here is that you could later define a
specialisation of std::list in which iterator was not a type. So the
compiler can't be certain that you actually intended for
std::list<T>::iterator to be a type until it knows the actual type of T
unless you explicitly say that std::list<T> is a type.

There are some genuine examples of where this ambiguity can cause
problems. You could probably do something like:

namespace std {
    template<> class list<MyOtherClass> {
        public:
            enum { iterator };
        ...

-- 
JDL

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