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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<ms[m]> hkaiser: this seems to fix the docs: https://github.com/STEllAR-GROUP/docker_build_env/pull/23
<ms[m]> they had a regression in breathe...
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<jbjnr> is my connection broken, or is nobody posting anything today?
<Yorlik> jbjnr: quack ;)
<zao> Simberg said something about the Docker env, otherwise nada.
<jbjnr> ok. Thanks - I see new messages. My riot window was stuck again and I reconnected.
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<weilewei> if I have a = (int) b * (uint64_t) c, will variable a becomes uint64_t automatically? Does it mean that lower rank type operation higher rank type guarantee to have a result of higher rank type?
<K-ballo> the variable won't change type.. is it an `auto a = ...;` kind of situation?
<weilewei> yes
<Yorlik> int can be negative, uint64_t not ...
<Yorlik> Interesting conflict
<K-ballo> the int argument will be promoted to uint64_t
<Yorlik> So - if it is -1 it will be 0xFFF...?
<weilewei> I see, thanks! K-ballo
<weilewei> Yorlik if b is negative, I doubt that promotion will be undefined value
<K-ballo> it is defined, used to be implementation-defined, now we have 2's complement so it would indeed be 0xffff...
<Yorlik> I wonder where there are type promotion rules written to look up. With the given example and statements bit width wins over signum.
<K-ballo> in modulo arithmetics -1 == 2^n -1
<weilewei> same question, where to read up this knowledge?
<Yorlik> Is that equivalent with 2-complement? I just recently heared that term "modulo arithmetics" for the first time, though I've been knowing % for ages already.
<K-ballo> weilewei: I'd recommend not knowing any of this
<K-ballo> the next guy coming up (maybe you in a few months) will be wondering what type `a` has
<K-ballo> modulo arithmetic is not equivalent to 2's complement
<weilewei> lol, really, but I really want to get my mind straight, I am dealing with large index which requires uint64_t, so I am trying to avoid type issues
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<weilewei> found this page, maybe a good read https://en.cppreference.com/w/cpp/language/implicit_conversion
<Yorlik> weilewei: Yup - that seems to be it.
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<weilewei> in `Integral conversions` section, it mentions about modulo arithmetic. The wording takes sometime to understand.. I will read it up
<K-ballo> it just means you take the module 2^n of the result (which is dropping higher bits)
<weilewei> K-ballo got it! thanks
<Yorlik> So x%(2^n) is basicvally a nice bit truncator, like and according bitwise and would be
<K-ballo> x%(2^n) === x & (2^n - 1) which is a mask with the first n bits on
<Yorlik> I wonder if there are some good resources out for bit arithmetic and all the things you can do with it. I recently had to do stuff with intrinsics and bit manibulations and realized my knowledge in that domain is horribly underdeveloped.
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<ms[m]> Yorlik: this is a pretty comprehensive resource on bit twiddling: https://graphics.stanford.edu/~seander/bithacks.html
<Yorlik> ms[m] Thanks alot good Sir ! :D
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<bita_> hkaiser, I am convinced that you were right about shape_d. I just added the pull request
<hkaiser> bita_: thanks a lot!
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<diehlpk_work> weilewei, yet?
<weilewei> diehlpk_work yes
<diehlpk_work> Check your Telegram please
<weilewei> say, uint64_t a; a = foo(b); where return type of foo is int. Will returned value of foo(b) be promoted to uint64_t?
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<weilewei> well, it will be overflowed if the returned value of foo(b) is larger than 2^31 - 1
<zao> A function with a particular return type always returns values of that type.
<zao> Promotions/conversions in the context in which it is called cannot percolate into the function.
<zao> The int, once returned, may be subject to conversions.
<weilewei> Right, I was expecting compiler will do some work there, but apparently I was naive
<zao> It's bound by the rules of the language, no amount of optimization effort can alter this :)
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<weilewei> Right...