Re: Problem with typename.

From:
"Victor Bazarov" <v.Abazarov@comAcast.net>
Newsgroups:
comp.lang.c++
Date:
Wed, 16 May 2007 16:02:19 -0400
Message-ID:
<f2fo0d$6c6$1@news.datemas.de>
StephQ wrote:

First of all: distinction of keywords typename and class in template
arguments.
Accoarding to a post in a well known moderated group:

"There are three possibilities for template arguments:

1) type as in "template <typename T>" or "template <class T>"
2) non-type as in "template <int N>"
3) class template as in "template <template<typename> class C>" "

Now, in 1 class and typename are equivalent.
1 is parametrizing the type of the object


1 is a beginning of a template declaration. WHAT it parameterises
(a type or a function) is defined _after_ the closing angle bracket.

template<typename T> class A; // declares a class template "A"
template<typename T> void foo(int); // declares a function template

2 is parametrizing which object


Uh... I don't understand that statement.

In 3 you have to use class, because C is just a template class.


Huh?

But I admit I am a little confused here, what is exactly C?


C is a template with exactly 1 argument which is itself a type.

Could I
pass stl vector as C ?


Mmm... No. 'std::vector' has more than 1 argument.

So template<typename> really means only that C
is a class template (and not a type), but actually could be defined by
more than 1 template argument? Where could I find a concrete easy
example?


On the web?

Also, what is considered a good programming syntax?


Any syntax your compiler doesn't barf at is "good programming syntax".

Using typename in
1 and 2, while class in 3?


Doesn't matter. Either will do.

Another problem I'm facing is that gcc requires the typename keyword
in a lot of situations, like:

template <class T>
void print(const std::vector<T>& vec)
{
 std::vector<T>::const_iterator i;
...
}

warning: `std::vector<T, std::allocator<_CharT> >::const_iterator' is
implicitly a typename (deprecated)

I don't know where should I add the typename keyword, and why.


To disambiguate anything as a type name. Usually it is required with
"dependent names". Look it up.

p.s. Is there a way to write a print function once for list, vector,
deque, .... ?


Yes, it's called "std::copy" or "std::transform" or "std::for_each".

    std::copy(mylist.begin(), mylist.end(),
                std::ostream_iterator<int>(std::cout, ", "));

V
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