Re: class inside of template puzzler

=?iso-8859-1?q?Daniel_Kr=FCgler?= <>
Sun, 23 Sep 2007 16:57:16 CST
On 22 Sep., 23:42, Jason Turner <> wrote:

#include <map>

I expect here one additional include:

#include <string>

template<class P1>
struct T
    struct T2
         T2() {}

       std::map<std::string, T2> m;

       std::map<std::string, T2>::iterator itr1 = m.begin();
         //error: expected `;' before 'itr'

       std::map<std::string, T::T2>::iterator itr2 = m.begin();
         //type/value mismatch at argument 2 in template parameter list
for 'template<class _Key, class _Tp, class _Compare, class _Alloc>
class std::map'
         // expected a type, got 'T<P1>::T2'
         // template argument 4 is invalid
         // expected initializer before 'itr2'

       std::map<std::string, T<P1>::T2 >::iterator itr3 = m.begin();
         //Same error as above

       std::map<std::string, T<int>::T2 >::iterator itr4 = m.begin();
         // Compiles, but not really useful

       typename std::map<std::string, T2>::iterator itr5 = m.begin();
         // Works as expected


int main() {}

I understand that typename is meant to be used in cases where the
compiler thinks there is an ambiguity and it does not know if you are
referring to a type or a variable. However, I don't understand why it
is is needed in the case above.

In all reclaimed cases it is indeed that the compiler cannot
know, whether

std::map<std::string, T2>::iterator

specifies a type or a non-type entity, because T2 is a
dependent type and therefore also std::map<std::string, T2>
is also type-depending. The compiler cannot foresee by the
basic language rules whether there exists a specialization
of std::map for *some special T2* which has e an iterator
that is a type. The problem would *not* exist, if T2 where
declared as completly known type, as e.g. in the following

struct TBase {
    struct T2
         T2() {}

template<class P1>
struct T : TBase
       std::map<std::string, T2> m;
       std::map<std::string, T2>::iterator itr1 = m.begin(); // OK
       std::map<std::string, typename T::T2>::iterator itr2 =
m.begin(); // OK
       std::map<std::string, typename T<P1>::T2 >::iterator itr3 =
m.begin(); // OK
       std::map<std::string, T<int>::T2 >::iterator itr4 =
m.begin(); // OK

Note that I added two typename prefixes in the
expressions that define itr2 and itr3, because in
both cause the second template argument of std::map
is still type-dependent (according to the currently
valid standard ISO-14882-2003, see [temp.res]/6).

Case in point, the declaration of "m" does not need "typename."

Yes and there is never a reason for that, because
the compiler requires to see <map>, which declares
the class template std::map. Since we don't have
overloading of class templates, it is clear, that
std::map<> must always be type - independent on
it's template arguments (This does of course not
guarantee, that for every template argument of
std::map every derived expression of std::map<T, ..>
is well-formed).

Greetings from Bremen,

Daniel Kr?gler

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