'mod' and -1 in haskell

Question

how come that following piece gives 4?

(\x -> (x-1) `mod` 5) 0

Yann Vernier · Answer 1 · 2010-07-25T10:59:32.003

6

Because mod is defined as positive. div and mod operate on a floor basis, while / and rem do not.

Prelude> let x=(-2)
Prelude> let y=5
Prelude> (x`div`y)*y+(x`mod`y)
-2
Prelude> y*(truncate ((fromInteger x)/fromInteger y)) + (x`rem`y)
-2
Prelude> x`rem`y
-2
Prelude> x`mod`y
3
Prelude> (fromInteger x)/fromInteger y
-0.4
Prelude> x`div`y
-1

Addendum: As KennyTM rightly points out, I should have used quot, not /:

Prelude> (x`quot`y)*y+(x`rem`y)
-2
Prelude> (x`quot`y)
0

I simply did not remember it, and was too hasty to look it up. quot will do an integer division.

edited Jul 25 '10 at 10:59

answered Jul 24 '10 at 11:37

Yann Vernier

15,414
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7

The correspondence of `div` should be `quot`, not `/`. – kennytm Jul 24 '10 at 12:12

score 5 · Accepted Answer · answered Jul 24 '10 at 11:25

5

-1 modulo 5 is, by definition, 4.

answered Jul 24 '10 at 11:25

IVlad

43,099
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it differs by programming language, and by which mod function you use – newacct Jul 24 '10 at 15:42

'mod' and -1 in haskell

2 Answers2