When should I use typename in a template function?

Question

According to this question, typename should be added to tell compiler that iterator is a type (Is it correct?), e.g.

template <typename T>
void print(vector<T> &v) {
  for (typename vector<T>::const_iterator it=v.begin(); it!=v.end(); ++it)
    cout<<*it<<endl;
}

Two questions:

Under what circumstances should a typename be added? Can someone give me another example?
Can auto in C++11 replace all these typename xxx cases?

This is one of the canonical examples of `auto` being useful. — chris, May 04 '14 at 02:11
1) See the link above. 2) Yes, this is one of the great benefits of `auto`, it replaces many of these annoying things. — Mikael Persson, May 04 '14 at 02:12
Of course if you want the proper type from `auto`, you'll need `cbegin`. — chris, May 04 '14 at 02:13

score 0 · Accepted Answer · edited May 04 '14 at 02:16

0

When you are referring to a typename whose exact meaning depends on a template parameter.
Not all, but many. auto has to be able to infer/deduce the type. In something like this:
```
typename T:foo_t var;
```

for example, it can't.

edited May 04 '14 at 02:16

David G

answered May 04 '14 at 02:14

kec

Thanks for the straightforward answer. I will take time to read the link above - quite a long post :) – Deqing May 04 '14 at 03:48
It's basically a hint to help the parser. – kec May 04 '14 at 04:03

1 Answers1