the integer part of the division

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hello everyone
how do i get the integer part of the output of a division i.e. 23/5=4 (and the remainder is 3)

Accepted Answer

Azzi Abdelmalek
Azzi Abdelmalek on 4 Oct 2012
Edited: Azzi Abdelmalek on 5 Oct 2012
fix(23/5) % integer part
rem(23,5) % remainder
%or
mod(23,5)
  3 Comments
Royi Avital
Royi Avital on 9 Apr 2021
It works only for Float64 and Float32.
For integer types use idivide.

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More Answers (1)

Walter Roberson
Walter Roberson on 5 Oct 2012
Consider -23/5 (that is, the case of negative numbers.) Should that be -4 times 5 and remainder -3, or should it be -5 times 5 and remainder 2 ?
If you want the -4 version so that abs() of the "integer part" of -23/5 and 23/5 are the same, then use fix()
If you want the -5 version so that the remainder is always non-negative, then use floor()
You can use either fix() or floor() if you only have positive values, with floor() being preferred (more efficient, mathematically clearer)
Do not use round() even if you have only positive values. round(23/5) is 5, not 4.

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