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Add exhaustive divrem tests for ZZ #1919

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Before I touch the implementations, I wanted to make sure to capture the current behavior.

Assumption: Base.divrem(x,y) == (Base.div(x,y), Base.rem(x,y) and Nemo.divrem(x,y) == (Nemo.div(x,y), Base.mod(x,y))

The few commented out tests are due to some weirdness with finite precision Ints together with unsiged Ints. All of the offending methods are in AA, and even in julia base are some discussions about what the return type of these should be.

One method I had to remove (there is a fallback via UInt -> ZZRingElem), because it did not satisfy the above assumption.

Does this look sensible?

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codecov bot commented Oct 23, 2024

Codecov Report

All modified and coverable lines are covered by tests ✅

Project coverage is 87.51%. Comparing base (3286337) to head (abcaa10).

Additional details and impacted files
@@            Coverage Diff             @@
##           master    #1919      +/-   ##
==========================================
+ Coverage   87.49%   87.51%   +0.02%     
==========================================
  Files          97       97              
  Lines       35537    35535       -2     
==========================================
+ Hits        31092    31099       +7     
+ Misses       4445     4436       -9     

☔ View full report in Codecov by Sentry.
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I think questions about such fundamental conventions are above my pay grade, sorry!

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fieker commented Oct 24, 2024 via email

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fieker commented Oct 24, 2024

... Maybe rename Nemo.divrem into divmod?

@@ -491,10 +491,6 @@ function rem(x::ZZRingElem, c::ZZRingElem)
return r
end

function rem(a::ZZRingElem, b::UInt)
return ccall((:fmpz_fdiv_ui, libflint), UInt, (Ref{ZZRingElem}, UInt), a, b)
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as this will remove ZZ(3) % UInt(2) I don't like it. Rather do an error if the input is illegal (ie. the remainder negative)

@lgoettgens lgoettgens marked this pull request as draft October 25, 2024 09:45
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3 participants