Subject: Re: [boost] [Review] Boost.Type Traits Extension by Frederic Bron
From: Frédéric Bron (frederic.bron_at_[hidden])
Date: 2011-03-18 02:05:43
> I've found some problems. In the ICL I have an
> operator +
>
> template <class Type>
> inline typename enable_if
> <is_associative_element_container<Type>, Type>::type
> operator + (Type object, const Type& operand);
>
> and is seems that more overloads are checked valid as actually exist.
> Is this intentional?
>
> BOOST_AUTO_TEST_CASE(has_op_extension_qualifiers)
> {
> typedef int T;
> typedef interval_set<T> IntervalSetT;
> typedef IntervalSetT::interval_type IntervalT;
>
> // This is supposed to succeed
> BOOST_CHECK((has_operator_plus<IntervalSetT, const IntervalSetT&,
> IntervalSetT>::value));
>
> BOOST_CHECK((!is_convertible<const IntervalSetT&, IntervalSetT&>::value));
>
> // These are supposed to fail, but they succeed
> BOOST_CHECK((has_operator_plus<IntervalSetT, IntervalSetT&,
> IntervalSetT>::value));
> BOOST_CHECK((has_operator_plus<IntervalSetT, IntervalSetT,
> IntervalSetT>::value));
> BOOST_CHECK((has_operator_plus<IntervalSetT, IntervalSetT,
> IntervalSetT const&>::value));
> }
Your question here is the same as the following example:
struct T { };
struct RET { };
RET f(T, const T&) { return RET(); }
void g(RET) { }
void h(const RET&) { }
int main() {
//Does this work?
{ T t1; const T &t2=t1; g(f(t1, t2)); } // -> yes
{ T t1; T &t2=t1; g(f(t1, t2)); } // -> yes
{ T t1; T t2; g(f(t1, t2)); } // -> yes
{ T t1; T t2; h(f(t1, t2)); } // -> yes
return 0;
}
What it means it that has_operator_plus<LHS, RHS, RET> does not check
for exact conformance of arguments LHS and RHS but checks if you can
use LHS and RHS as arguments to the operator. And clearly you can use
a lot of types when your operator is declared taking a const RHS&.
Hope this helps. I think this policy is desirable otherwise you would
have to test many options, especially with binary operators before
being sure you can use the operator.
Frédéric