CODEWITHSUNDEEP
haskell
Haskell
Haskell

Reputation: 125

Haskell. How this function catch negative numbers as a parameter?

I don't understand, how this function negative numbers catchs? 

mdrop 0 []    = []
mdrop 0 (h:t) = h:t
mdrop n []    = []
mdrop n (h:t) = mdrop (n-1) t

If my input looks like this mdrop (-3) [1, 2, 3, 4, 5, 7, 88, 6] I receive an empty list. But I don't understand in which line catches the negative number.

Upvotes: 2

Views: 1277

Answers (1)

chepner
chepner

Reputation: 531490

Nothing catches the negative number specifically, but decreasing n is always accompanied by a shorter list, so eventually you hit the empty list and the 3rd equation matches. n is an example of an irrefutable pattern; it will match any number (positive, zero, or negative), although with the given definition, it will never be checked until you've already confirmed that the first argument is non-zero.

Effectively, a negative argument produces the same result as a sufficiently large argument, in that it can always be decremented without reaching 0.

Tracing it manually,

     mdrop (-3) [1,2,3,4,5,7,88,6] 
  == mdrop (-4) [2,3,4,5,7,88,6]
  == mdrop (-5) [3,4,5,7,88,6]
  == mdrop (-6) [4,5,7,88,6]
  == mdrop (-7) [5,7,88,6]
  == mdrop (-8) [7,88,6]
  == mdrop (-9) [88,6]
  == mdrop (-10) [6]
  == mdrop (-11) []
  == []

As an aside, your first two definitions could be replaced by the single definition

mdrop 0 xs = xs

since in both original definitions, you simply return the second argument when you the first matches 0.

Upvotes: 7

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