Issue #12508 has been updated by Yukihiro Matsumoto.

Status changed from Open to Feedback

Instead, I propose `pow(a)` and `pow(a,b)` where the latter works as `mod_pow()` here.

Matz.


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Feature #12508: Integer#mod_pow
https://bugs.ruby-lang.org/issues/12508#change-60008

* Author: Makoto Kishimoto
* Status: Feedback
* Priority: Normal
* Assignee: 
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A new method Integer#mod_pow, power with modulo.

a.mod_pow(b, m)  #=>  (a**b) % m

Sometimes a**b becomes very large number, then naive
implementation may be unefficient. Fast implementation
is useful.
(with USE_GMP symbol, this implement uses mpz_powm() )

(see https://github.com/ruby/ruby/pull/1320 )



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