Hi Steve, Concepts are able to define default implementations if the type doesn't provides the specific signature. For example in 20.1.2 Comparisons the concept LessThanComparable defines a default for implementation for bool operator>(U const& a, T const& b) { return b < a; } bool operator<=(U const& a, T const& b) { return !(b < a); } bool operator>=(T const& a, U const& b) { return !(a < b); } In the definition of the concept less_than_comparable in Boost.Any, you define the implementation of these operators in function of operator<(), but if the types provide the specific signature the specific function is not called. I guess that the default implementation of for example operator<= should use operator< only if the the underlying types don't provide operator<=. What do you think? Best, Vicente -- View this message in context: http://boost.2283326.n4.nabble.com/type-erasure-default-implementation-of-co... Sent from the Boost - Dev mailing list archive at Nabble.com.