AMDG Frédéric Bron wrote:
Here, I cannot find a way to get the return type of operator<(T, U). The only thing I can do is:
1. do not treat "void operator<(T, U)" => thus using is_less_comparable<T, U> if operator<(T, U) returns void will be a compilation error => no compiler warning for the ternary operator, trick 2. treat "void operator<(T, U)" => must use the ternary operator, trick => compiler warning => is_less_comparable<T, U> is true_type if operator<(T, U) exists and returns void (no compiler error)
I am in favour of solution 1 because when we need is_less_comparable<T, U> it is obviously to later do something like "bool result=t<u". Then with solution 1 the compilation error would come when is_less_comparable<T, U> is invoked and with solution 2 it would come when t<u is invoked. So in either case, there will be a compilation error somewhere.
Other issue: I do not see how I can test the return type of operator< to check if it is convertible to bool.
Once you know that operator< exists and returns non-void, you can use simple overload resolution. typedef char no; struct yes { no dummy[2]; }; yes is_bool(bool); no is_bool(...); sizeof(is_bool(T() < U())) == sizeof(yes); In Christ, Steven Watanabe