Member template function. Why doesn't this compile?

Question

I whant to iterate in a member function over a list. In this code sample I am getting compiler error in for-line:

Expected ; before it, it not declared, end not declared.

Why? The function is even not used in code!

template <class T> bool Settings::saveSimpleList( QString k, const T & l ){
    //...
    for ( T::ConstIterator it = l.constBegin(), end =l.constEnd(); it != end; ++it )
    {
        QString itemValue = QVariant( *it ).toString();
        //...
    }

    return true;
}

I see, I am missing something in template programming. Thank you for your hints!

BTW: I voted to reject your proposed edit of adding the `qt` tag. The essence of the question doesn't have any relation to QT. — Jon, Mar 05 '11 at 13:20
Check out this thread : http://stackoverflow.com/questions/610245/where-to-put-the-template-and-typename-on-dependent-names — Prasoon Saurav, Mar 05 '11 at 17:20

score 7 · Accepted Answer · answered Mar 05 '11 at 13:14

7

Do this:

for (typename T::ConstIterator it = l.constBegin(), end =l.constEnd(); it != end; ++it )
{
    QString itemValue = QVariant( *it ).toString();
    //...
}

When writing T::ConstIterator, the compiler can either interpret that as "the static member ConstIterator of type T" or as "the type defined as ConstIterator by a typedef in type T".

If your intent is the second, as here, you need to add typename to tell the compiler.

answered Mar 05 '11 at 13:14

Jon

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+1: Thank you!!! I've read about typename in this context, but this is the first time I've using it. – Valentin H Mar 05 '11 at 13:18

score 1 · Answer 2 · answered Mar 05 '11 at 13:15

1

The compiler cannot know that T::ConstIterator is a type (because it does't know what T will be). You can indicate that by adding a typename in front of it.

answered Mar 05 '11 at 13:15

Bo Persson

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Member template function. Why doesn't this compile?

2 Answers2

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