hkaiser changed the topic of #ste||ar to: STE||AR: Systems Technology, Emergent Parallelism, and Algorithm Research | stellar.cct.lsu.edu | HPX: A cure for performance impaired parallel applications | github.com/STEllAR-GROUP/hpx | Buildbot: http://rostam.cct.lsu.edu/ | Log: http://irclog.cct.lsu.edu/ | GSoC: https://github.com/STEllAR-GROUP/hpx/wiki/Google-Summer-of-Code-%28GSoC%29-2020
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<sidhu> Guys how can I keep myself signed in or remembered on this channel? I had to join again now all previous conversations are gone
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<sidhu> I had to join again'=(
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<sidhu> someone please help me, do i always need to join again?
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<7F1AAI3TQ> You could use the matrix channel. IRC itself does not do backlogging
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<sidhu> okay
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<sidhu> is this organization's chat available on matrix?
<K-ballo> it is
<sidhu> can i have the link or group name etc anything like that so that can join
<K-ballo> I can't find it, but some of the people here with [m] in their names should know
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<K-ballo> seems you found it
<sidhu[m]> yes i just did
<K-ballo> where did you find it?
<sidhu[m]> tried different name for searching
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<sidhu[m]> room name is hpx
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<nikunj> sidhu[m], thanks for the room. I guess I'll bid hexchat farewell.
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<zao> All of freenode is accessible via a server-level bridge.
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<gnikunj[m]> @hka
<gnikunj[m]> hkaiser: I thought you said, no more commits yesterday :/
<hkaiser> gnikunj[m]: found some problems, still
<hkaiser> looks good now, I believe
<gnikunj[m]> hkaiser: I'll believe you
<zao> No more (known) bugs? :P
<hkaiser> there are no bugs
<hkaiser> just technology gaps ;-)
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<gonidelis[m]> hkaiser: yt?
<gonidelis[m]> Does anyone know what is the difference between the two specializations of `in_out_result` here
<gonidelis[m]> ?
<gonidelis[m]> Actually does that `constexpr operator in_out_result<>` overloads the `in_out_result()`? And if yes, what's the difference between having an `const & ` and a `&&` after that?
<zao> gonidelis[m]: I had never seen "ref-qualified member functions" before, heh... https://en.cppreference.com/w/cpp/language/member_functions
<zao> In particular, the distinction between no ref-qualifier and lvalue ref-qualifier may be helpful in understanding the meaning.
<gonidelis[m]> no ref-qualified means that I cannot write `&foo()` ?
<zao> It's about what kinds of objects it can bind to when called, and affects overload resolution.
<zao> It lets you overload on the lvalue/rvalue-ness of the object it's being called on.
<zao> (I don't know these newfangled C++ features, I just google :P )
<K-ballo> it's essentially a "copy" conversion operator and a "move" conversion operator
<K-ballo> imagine those as being converting constructors and it should become obvious
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<hkaiser> gonidelis[m]: it's what K-ballo said, just that those conversions are defined on the source type instead of the destination type (as for converting copy/move constructors)
<gonidelis[m]> What do we need these "copy" and "move" conversions?
<gonidelis[m]> plus
<gonidelis[m]> Why do we need two constructors (tha behave like that)?
<gonidelis[m]> (maybe that's the same question)
<gonidelis[m]> K-ballo: hkaiser
<hkaiser> gonidelis[m]: try to understand the difference between copying and moving an object (in C++)
<hkaiser> that will answer this question
<gonidelis[m]> ok will do. and where could I look for these syntaxes ` constexpr operator in_out_result<I2, O2>() &&
<gonidelis[m]> `
<gonidelis[m]> ` constexpr operator in_out_result<I2, O2>() const &`?
<hkaiser> the best explanation of move semantics I found is Howard Hinnant's talk about it back in the early 2000's (2004?)
<hkaiser> struct Bla {operator Foo() {}}; is used by the compiler whenever an instance of a Bla is used in place of an instance of a Foo
<hkaiser> the same is true for struct Foo { Foo(Bar) {...}};
<gonidelis[m]> hkaiser: ok thanks a lot... I think I should refresh your lectures first ;)
<hkaiser> the operator defines the conversion operation on the source type (Bar), while the converting constructor defines it on the destination type (Foo)
<hkaiser> both express the information of how to convert a Bar into a Foo