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Why doesn't this return type work? (C++)

Answer 1

+5 A:

iterator is not yet known at that point. You need to tell it it's in the list<T> class:

template<class T>
typename list<T>::iterator& list<T>::iterator::operator++() {
    (this->lstptr)->current = ((this->lstptr)->current)->next;
    return *this; // *this here, since this is a pointer only
}

Notice the typename is required, since list<T>::iterator is a type prefixed with a template specialization, and you need to tell the compiler about that - in spite of the fact that Visual C++ probably will accept code not putting typename before it. Omitting the typename, the compiler should assume it's not a type and should sort of produce the same error message.

You could safe yourself that hassle by putting the code straight into the class:

template<class T>
class list<T>::iterator
{
public:
iterator& operator++() {
    (this->lstptr)->current = ((this->lstptr)->current)->next;
    return *this; // *this here, since this is a pointer only
}
protected:
    list* lstptr;
};

Johannes Schaub - litb 2009-05-21 01:53:31

Keand64 2009-05-21 01:57:26

Oh nice. Seems like the compiler does require it at some places actually. In some posts i've seen people using msvc++ are not required to put it at certain places where it is needed normally :)

Johannes Schaub - litb 2009-05-21 02:00:54

Answer 2

+1 A:

litb has answered your question completely. I think it worth highlighting that in an effort to make C++ easier to use, the C++ Committee has added a new syntax for the declaration of functions. The result is that you'll be able to define your function as follows (n2541) without needing the extra qualification:

template<class T>
auto list<T>::iterator::operator++()->iterator& 
{
  // ...
}

According to the supported feature list, GCC 4.4 already has this feature.

Richard Corden 2009-05-21 08:47:59

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Why doesn't this return type work? (C++)

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