Leonidas changed the topic of #ocaml to: Discussion about the OCaml programming language | http://www.ocaml.org | OCaml 5.1.1 released: https://ocaml.org/releases/5.1.1 | Try OCaml in your browser: https://try.ocamlpro.com | Public channel logs at https://libera.irclog.whitequark.org/ocaml/
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<discocaml> <jobhdez> Are the llvm bindings maintained and will continue to be maintained?
<discocaml> <jobhdez> Are they current?
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<discocaml> <sim642> I think there's enough community interest behind it: https://discuss.ocaml.org/t/ann-llvm-15-is-out/13019. That at least claims OCaml Software Foundation support for maintenance.
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<ania123>  If we know n = N div 2^k and 2^k * (n mod 2) + N mod 2^k can we conclude N mod 2^(k+1)?
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<discocaml> <contextfreebeer> ania123: did you type that in correctly? personally I am unable to understand exactly what you're asking
<discocaml> <contextfreebeer> I think they aren't online anymore actually
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<discocaml> <yawaramin> seems like asked in wrong channel by accident
<discocaml> <Et7f3 (@me on reply)> You have both assertions:
<discocaml> <Et7f3 (@me on reply)> n = N div 2^k
<discocaml> <Et7f3 (@me on reply)> n = (2^k) * (n mod 2) + N mod 2^k
<discocaml> <Et7f3 (@me on reply)> And you want to conclude:
<discocaml> <Et7f3 (@me on reply)> N mod 2^(k+1)
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<discocaml> <Et7f3 (@me on reply)> I wasn't able to prove: might be I haven't tried hard enough or the question is bad typed.
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