Subject: Re: [boost] [type_traits] adding is_less_comparable<T>, etc.
From: FrÃ©dÃ©ric Bron (frederic.bron_at_[hidden])
Date: 2009-09-01 09:01:24
> remove_cv.html is not in the archive (the link from index.html was broken).
It is already part of boost but as I use it I should include it in the
> The links for the additions are not in alphabetical order on index.html.
Yes, I thought it was better here to have is_less close to
is_less_equal but I have no strong opinion on that and can come back
to alphabetical order.
> Even if you didn't invent the code, you assembled and documented it, so your name should be in the copyright notice (in the documentation), right? Â Add others to the list as you think appropriate.
OK, I added Roman Perepelitsa and my-self. Roman, if you are still
reading this mailing list, do you agree to have your name in the
> Your HTML formatting differs from the original, though that could be the means by which it was generated rather than a difference in the formatting of your files.
I think so because I applied the method recommended by Beman.
> Your content will appear in a single page, reference.html, if added to Boost.TypeTraits, so why create separate pages now?
I do not understand your point here. Do you mean I could have all
traits on the same page and only one line in reference.html?
> The description of is_greater_comparable should be more specific:
> Â If a value of type T can be compared with another using
> Â operator > to determine whether the first is greater than
> Â the other, then inherits from true_type, otherwise
> Â inherits from false_type.
OK for that.
> I think is_greater_than can be less restrictive than it is. Â std::greater() provides an operator that permits comparing unrelated types as T's. Â Therefore, I wonder if this would work:
> Â template <class T, class U = T>
> Â struct is_greater_comparable;
> Then, is_greater_comparable<int, float> would inherit from true_type.
Thank you for suggestion (applied) and help finding my error (shame on me).
>> I can add that one important question of the review is: should we
>> report that T is less comparable if T<T returns void?
> I think that is not appropriate as the whole point of less-than-comparable is to be able to determine whether one instance is less than another, not that "operator <" is supported.
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.
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