Re: enforcing const overload of a method
On 07/04/2010 09:42 AM, Rem wrote:
Please take a look at this code:
class A
{
public:
template<class XIter>
A(const XIter begin, const XIter beyond)
{}
};
#include <vector>
int main(int argc, char* argv[])
{
std::vector<int> integers;
A a1( integers.begin(), integers.end() );//No problem
std::vector<int>::const_iterator i( ++( integers.begin() ) );//
compiler doesn't like const_iterator here
A a2( i, integers.end() );//error C2660: 'A::A' : function does not
take 2 arguments
return 0;
}
What happened here is that in a2 constructor call Visual C++ 2008 was
unable to choose the const version of std::vector<T>::end() - or so I
guess. The error message is very confusing.
Do you know how I can make sure that the const_iterator is picked as a
template argument instead of iterator?
Personally I would never create a template definition like that. But a
simple solution to your problem would be the use of a const reference
std::vector<int> integers;
const std::vector<int> &r = integers;
A a1( integers.begin(), integers.end() );//No problem
// compiler doesn't like const_iterator here
std::vector<int>::const_iterator i( ++( integers.begin() ) );
// will work as r.end() will return a const_iterator
A a2( i, r.end() );
return 0;
HTH
cpp4ever
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