Reputation: 4286
I'm trying to return an iterator for a vector in a templated class (I'm not sure if that makes a difference, but I've read that may, so I thought I'd mention it). The problem is that I get an error about C++ not supporting default-int when I try this. I've looked online and from what I can see in forums and explanaions, I don't think I'm that far off, it just won't compile.
template<class T>
class Table
{
public:
...
vector<shared_ptr<vector<T>>>::iterator GetRowIterator();
//vector<shared_ptr<vector<CellValueType> > >::const_iterator GetRowIterator();
...
protected:
vector<shared_ptr<vector<CellValueType> > > data; //outside vector is rows, inside vector is columns
...
};
vector<shared_ptr<vector<T> > >::const_iterator Table<T>::GetRowIterator()
{
return data.begin();
}
The errors that I get are:
error C2146: syntax error : missing ';' before identifier 'GetRowIterator'
error C4430: missing type specifier - int assumed. Note: C++ does not support default-int
Edit:
Changed the end angle brackets so they are not as close together - same error.
Any thoughts as to why this is occurring?
As always, thanks for advice/help in advance!
Upvotes: 1
Views: 1645
Reputation: 16039
Also remember to use typename when declaring the template-dependent return type:
typename vector< shared_ptr< vector< T > > >::iterator GetRowIterator();
and the method definition
typename vector< shared_ptr< vector< T > > >::const_iterator Table<T>::GetRowIterator()
{
return data.begin();
}
Notice also that when defining a template class method outside the class definition, you have to use the template keyword:
template <class T> typename vector< shared_ptr< vector< T > > >::const_iterator Table<T>::GetRowIterator()
{
return data.begin();
}
So that the compiler can know what the T is about.
Upvotes: 4
Reputation: 264361
This part here:
vector<shared_ptr<vector<T>>>
It is a problem with the C++ syntax you can not put >> together like that.
vector<shared_ptr<vector<T> > >
This is a problem that is being addressed by the new standard.
Because the lexer is the first stage of the compiler it sees the >>> as a shift left operator followed by >. Thus you are getting syntax errors in your code. To supress this problem you just need to add white space between the > when closing templates.
Upvotes: 3