Hi,
the following templated code doesn't compile. It gives the error:
template_it.cc: 17: error: type 'std::vector<De rived1<T>*,
std::allocator< Derived1<T>*' is not derived from type 'Derived2<T>'
template_it.cc: 17: error: expected ';' before 'vi'
However, the same code compiles fine when I don't use any templates.
Also, why ist it okay to define (in class Derived2) the vector v but not
okay to define the iterator?
Thanks,
Ralf
#include <vector>
using namespace std;
template <class ID_TYPEclass Base
{
int id;
};
template <class Tclass Derived1: public Base<T>
{
};
template <class Tclass Derived2: public Base<T>
{
vector<Derived1 <T>*v;
vector<Derived1 <T>*>::iterat or vi;
};
int main()
{
return 0;
}
7  2683 
Dear Ralf,
Ralf Goertz wrote:
Hi,
the following templated code doesn't compile. It gives the error:
template_it.cc: 17: error: type 'std::vector<De rived1<T>*,
std::allocator< Derived1<T>*' is not derived from type 'Derived2<T>'
template_it.cc: 17: error: expected ';' before 'vi'
because the code is ambiguous.
template <class Tclass Derived2: public Base<T>
{
vector<Derived1 <T>*v;
vector<Derived1 <T>*>::iterat or vi;
};
change to:
template <class Tclass Derived2: public Base<T>
{
vector<Derived1 <T>*v;
typename vector<Derived1 <T>*>::iterat or vi;
};
Best regards,
Zeppe
On Aug 5, 5:07 am, Ralf Goertz
<r_goe...@expir es-2006-11-30.arcornews.de wrote:
Hi,
the following templated code doesn't compile. It gives the error:
template_it.cc: 17: error: type 'std::vector<De rived1<T>*,
std::allocator< Derived1<T>*' is not derived from type 'Derived2<T>'
template_it.cc: 17: error: expected ';' before 'vi'
However, the same code compiles fine when I don't use any templates.
Also, why ist it okay to define (in class Derived2) the vector v but not
okay to define the iterator?
Thanks,
Ralf
#include <vector>
using namespace std;
template <class ID_TYPEclass Base
{
int id;
};
template <class Tclass Derived1: public Base<T>
{
};
template <class Tclass Derived2: public Base<T>
{
vector<Derived1 <T>*v;
typename std::vector< Derived1<T>* v;
vector<Derived1 <T>*>::iterat or vi;
same as above
a) What are dependant types?
b) keeping an iterator to a vector as a member here is bad news
if that vector resizes
c) why are you storing pointers? What is RAII?
>
};
int main()
{
return 0;
}
Salt_Peter wrote:
On Aug 5, 5:07 am, Ralf Goertz
>>
template <class Tclass Derived2: public Base<T>
{
vector<Derived1 <T>*v;
typename std::vector< Derived1<T>* v;
> vector<Derived1 <T>*>::iterat or vi;
same as above
a) What are dependant types?
Derived1 and Derived2 are two kinds of similar objects. I have two large
vectors, one of each kind. Each object of the Derived2 vector is
dependent on very many (but not all) objects in the Derived1 vector.
b) keeping an iterator to a vector as a member here is bad news
if that vector resizes c) why are you storing pointers? What is
RAII?
I don't keep the iterator. I just used it here for illustration. I use
pointers since Derived1 is quite large and memory is an issue.
Furthermore, both vectors are filled via push_back and then they are not
touched anymore (only the objects inside the vectors are modified). So I
think I'm safe using pointers.
Ralf
Ralf Goertz wrote:
Salt_Peter wrote:
>On Aug 5, 5:07 am, Ralf Goertz
>>>
template <class Tclass Derived2: public Base<T>
{
vector<Derived1 <T>*v;
typename std::vector< Derived1<T>* v;
BTW, this gives an error: "expected nested-name-specifier" using g++
version 4.2.1
template <class Tclass Derived2: public Base<T>
{
* * *vector<Derived 1<T>*v;
* * *typename vector<Derived1 <T>*>::iterat or vi;
};
Why do we need a typename here? I don't see how it can resolve to
something else.
Thanks
puzzlecracker wrote:
>template <class Tclass Derived2: public Base<T>
{
vector<Derived 1<T>*v;
typename vector<Derived1 <T>*>::iterat or vi;
};
Why do we need a typename here? I don't see how it can resolve to
something else.
The compiler doesn't have any special treatment for std::vector. In
general, there could be a specialization for some Ts with a different
definition.
template<class T>
class weird
{
class iterator;
};
template<>
class weird<int>
{
int iterator;
};
Now what is iterator here?
template<class T>
class use
{
typedef weird<T>::itera tor fails_sometimes ;
};
Bo Persson
Bo Persson wrote:
puzzlecracker wrote:
>>template <class Tclass Derived2: public Base<T>
{
vector<Derive d1<T>*v;
typename vector<Derived1 <T>*>::iterat or vi;
};
Why do we need a typename here? I don't see how it can resolve to
something else.
The compiler doesn't have any special treatment for std::vector. In
general, there could be a specialization for some Ts with a
different definition.
template<class T>
class weird
{
class iterator;
};
template<>
class weird<int>
{
int iterator;
};
Now what is iterator here?
template<class T>
class use
{
typedef weird<T>::itera tor fails_sometimes ;
};
Fails always of course, as it should be
typedef typename weird<T>::itera tor fails_sometimes ;
See how easy it is to miss this ! :-)
Bo Persson This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics | |
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