
On Thu, Feb 15, 2018 at 10:11 AM, Steven Watanabe via Boost < boost@lists.boost.org> wrote:
AMDG
On 02/14/2018 10:29 PM, Zach Laine via Boost wrote:
On Wed, Feb 14, 2018 at 4:45 PM, Steven Watanabe via Boost < boost@lists.boost.org> wrote:
On 02/14/2018 03:11 PM, Zach Laine via Boost wrote:
On Tue, Feb 13, 2018 at 4:51 PM, Steven Watanabe via Boost < boost@lists.boost.org> wrote:
algorithm.hpp:
221: template <long long I, typename Expr> decltype(auto) get (Expr && expr, hana::llong<I> i); Does this need to be restricted to hana::llong, rather than, say, std::integral_constant?
It's the lack of nice syntax for integral_constant literals that made me choose this. I write get(expr, N_c) a lot, and I expect users to as well.
Supporting an IntegralConstant doesn't mean that you can't pass 0_c.
Is the change from std::integral_constant to IntegralConstant significant? That is, are you asking that I accept models of some concept, or just std::integral_constant?
I meant hana::IntegralConstant in the first place. std::integral_constant was just an example of another IntegralConstant, which hana::at can accept.
Great. Will do then.
<snip>
expression_free_operators.hpp/expression_if_else.hpp:
- Why is this not part of expression.hpp?
For the same reason that expression_free_operators.hpp is not. Those three headers are separate pieces that you may want some or all of, so I kept them separate.
expression_if_else makes a certain amount of sense as being separate. Making expression_free_operators separate comes with a high risk of ODR problems in any code that has different behavior depending on whether a given operator is defined. It's extra strange because expression_free_operators.hpp only has half the operator overloads, with the other half being defined as members.
I've read this a few times and don't get it. Without including expression_free_operators.hpp, where might the other ODR-colliding operators come from? Do you mean if the user defines her own operators
for
expression<>, and then includes expression_free_operators.hpp in another TU, or something else?
I was talking about the one definition rule as applied to some function that uses the operators. As an example, consider something that you yourself have written:
boost/yap/print.hpp:
template <typename T, typename = void_t<>> struct printer { std::ostream & operator() (std::ostream & os, T const &) { return os << "<<unprintable-value>>"; } };
template <typename T> struct printer< T, void_t<decltype( std::declval<std::ostream &>() << std::declval<T const &>() )> > { std::ostream & operator() (std::ostream & os, T const & x) { return os << x; } };
Now consider a class that defines a stream operator separately from the class:
// x.hpp class X {};
// x_print.hpp std::ostream& operator<<(std::ostream&, const X&);
Next, I have two separate translation units:
//a.cpp #include <x.hpp> #include <x_print.hpp> yap::detail::print_value(cout, X{});
// b.cpp #include <x.hpp> yap::detail::print_value(cout, X{});
printer chooses a different specialization in a.cpp than it does in b.cpp, which results in undefined behavior.
As a result of this problem, I believe that any free functions (including operators) that are intended to be found by ADL *must* be declared alongside the class definition.
Ah, thanks. I'll probably just jam everything into expression.hpp then. Zach