Re: enforcing const overload of a method

From:

Pete Becker <pete@versatilecoding.com>

Newsgroups:

comp.lang.c++.moderated

Date:

Sun, 4 Jul 2010 15:57:10 CST

Message-ID:

<2010070411291350813-pete@versatilecodingcom>

On 2010-07-03 22:42:26 -0400, Rem said:

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.

Don't guess. Learn the rules. The template constructor takes two arguments of the same type. The code passes two different types: the first argument has type const_iterator and the second is iterator.

The choice of an overloaded function depends on the arguments, not on how the result is used. The compiler picks the non-const version of end() because the integers object is not const.

The old way of getting around this was static_cast<const std::vector<T>&>(integers).end(). The new way is integers.cend().

--
Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The Standard C++ Library Extensions: a Tutorial and Reference (www.petebecker.com/tr1book)

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