counting recursive calls - Haskell

Question

Is there a function that count recursive calls for Extended Euclidean algorithm function?

rechnerb :: Integer -> Integer -> (Integer,Integer,Integer)
rechnerb 0 b = (b, 0, 1)
rechnerb a b = let (g, x, y) = rechnerb (b `mod` a) a
           in (g, y - (b `div` a) * x, x)

It would likely break [referential transparency](https://stackoverflow.com/a/9859966/67579) by counting recursive calls. — Willem Van Onsem, Dec 12 '19 at 13:05
The simplest makeshift way is to pass another parameter to your function like `cnt` which gets incremented by one every time you make a recursive call and include it in the return value somehow.. perhaps use a tuple for the return value whereas `fst` is the answer and `snd` is the `cnt`. Kind of... — Redu, Dec 12 '19 at 13:27
can you please write the program, I didn't really understand :/ @Redu — Goob99, Dec 12 '19 at 14:03

score 2 · Accepted Answer · answered Dec 12 '19 at 14:39

You may do like

rechnerb :: Integer -> Integer -> Int -> ((Integer,Integer,Integer), Int)
rechnerb 0 b c = ((b, 0, 1), c)
rechnerb a b c = let ((g, x, y), cnt) = rechnerb (b `mod` a) a (c+1)
                 in ((g, y - (b `div` a) * x, x), cnt)

and call like;

λ> rechnerb 7 1100 1 -- last parameter is the initial value (call # 1)
((1,-157,1),3)       -- (1,-157,1) is the answer and 3 is the # of calls to rechnerb

counting recursive calls - Haskell

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