Click here to Skip to main content
15,848,391 members
Please Sign up or sign in to vote.
4.00/5 (1 vote)
See more:
I created a template class that has a deque object. I infrequently want to be able access the std::deque::iterator object. However, I have been unable to get a very simple example to work... Any ideas how to fix it?
I have an example working now using protected and private inheritance, but I would prefer layering for performance reasons and generality if my class has more than one deque/vector. Moreover, with layering, I had hide more implementation details...

1st try

template<typename T>
class XYZ {
std::deque<T> B;
....
public:
std::deque<T>::iterator begin() { return B.begin(); }
);


warning C4346: 'std::deque<Task>::iterator' : dependent name is not a type
error C2146: syntax error : missing ';' before identifier 'begin'

2nd try

template<typename T>
class XYZ {
std::deque<T> B;
....
public:
std::_Deque_iterator<T, std::allocator< std::deque<T> > > begin() { return B.begin(); }
};


Strange error message:

error C2440: 'initializing' : cannot convert from 'std::_Deque_iterator<_Ty,_Alloc>' to 'std::_Deque_iterator<_Ty,_Alloc>'
Posted
Updated 13-Dec-10 20:57pm
v7
Comments
JF2015 14-Dec-10 1:34am    
Edited to add code formatting.

1 solution

Hi Ted,
Not really understanding why you want to expose that, anyhow this should compile:
C++
typename std::deque<T>::iterator begin()
{
    return B.begin();
}

cheers,
AR
 
Share this answer
 

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)



CodeProject, 20 Bay Street, 11th Floor Toronto, Ontario, Canada M5J 2N8 +1 (416) 849-8900