Ambiguous call to templated function due to ADL

Question

I've been bitten by this problem a couple of times and so have my colleagues. When compiling

#include <deque>
#include <boost/algorithm/string/find.hpp>
#include <boost/operators.hpp>

template< class Rng, class T >    
typename boost::range_iterator<Rng>::type find( Rng& rng, T const& t ) {
      return std::find( boost::begin(rng), boost::end(rng), t );
}

struct STest {
      bool operator==(STest const& test) const { return true; }
};

struct STest2 : boost::equality_comparable<STest2>   {
      bool operator==(STest2 const& test) const { return true; }
};

void main() {
      std::deque<STest> deq;
      find( deq, STest() ); // works
      find( deq, STest2() ); // C2668: 'find' : ambiguous call to overloaded function
}

...the VS9 compiler fails when compiling the second find. This is due to the fact that STest2 inherits from a type that is defined in boost namespace which triggers the compiler to try ADL which finds boost::algorithm::find(RangeT& Input, const FinderT& Finder).

An obvious solution is to prefix the call to find(…) with "::" but why is this necessary? There is a perfectly valid match in the global namespace, so why invoke Argument-Dependent Lookup? Can anybody explain the rationale here?

Answer 1

ADL isn't a fallback mechanism to use when "normal" overload resolution fails, functions found by ADL are just as viable as functions found by normal lookup.

If ADL was a fallback solution then you might easily fall into the trap were a function was used even when there was another function that was a better match but only visible via ADL. This would seem especially strange in the case of (for example) operator overloads. You wouldn't want two objects to be compared via an operator== for types that they could be implicitly converted to when there exists a perfectly good operator== in the appropriate namespace.