Could Not Match Expected Type Integer with actual type float

Question

countSequences :: Int -> Int -> Integer

countSequences 0 m = 0
countSequences m 0 = 0
countSequences (n) (m) =  if (n <= (m+1)) then (truncate((cee (m+1) (n) (0))) + truncate((countSequences (fromIntegral (n-1)) (fromIntegral (m))))) 
else truncate(countSequences (fromIntegral (n-1)) (fromIntegral (m)))

factorial :: Float -> Float
factorial 0 = 1
factorial 1 = 1
factorial x = x * factorial(x-1)

cee :: Float -> Float -> Float -> Float

cee x y z = if (x==y) then ((1) / (factorial ((x+z)-(y)))) else ((x) * (cee (x-1) (y) (z+1)))

i cant really understand why this error keep coming up .. the truncate is supposed to convert the type from Float to Integer so ..

opqdonut · Accepted Answer · 2011-12-20 22:02:50Z

5

The error is:

Couldn't match expected type `Float' with actual type `Int'
In the first argument of `(+)', namely `m'
In the first argument of `cee', namely `(m + 1)'
In the first argument of `truncate', namely
  `((cee (m + 1) (n) (0)))'

You see, the problem is you passing an Int to the function cee.

Here, I cleaned up the code for you:

countSequences :: Int -> Int -> Integer

countSequences 0 m = 0
countSequences m 0 = 0
countSequences n m = 
  if n <= m+1
  then truncate (cee (fromIntegral (m+1)) (fromIntegral n) 0) +
       countSequences (n-1) m
  else countSequences (n-1) m

factorial :: Float -> Float
factorial 0 = 1
factorial 1 = 1
factorial x = x * factorial (x-1)

cee :: Float -> Float -> Float -> Float

cee x y z =
  if (x==y)
  then 1 / factorial (x+z-y)
  else x * cee (x-1) y (z+1)

answered Dec 20, 2011 at 22:02

opqdonut

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3 Comments

ehird Over a year ago

fromIntegral m+1 should probably be fromIntegral m + 1 or fromIntegral (m+1), since it's such a common mistake to not realise that application binds tightest :)

opqdonut Over a year ago

@ehird: you're right. I got lazy and just used the convenient fact that (fromIntegral x) + 1 == fromIntegral (x+1). I'll edit the answer.

ehird Over a year ago

Technically, for sufficiently pathological x you could get an overflow pre-conversion, which would probably result in different results since Float doesn't overflow in the same way... but yes, close enough for all reasonable values :)

Stefan Holdermans · Accepted Answer · 2011-12-20 22:13:04Z

2

m is of type Int (per your type signature for countSequences: hence, so is m + 1. However, your function cee expects a Float, so the type checker righteously complains.

Furthermore, you will need a couple of more fixes to make this type check. Here's a version that passes the checker:

countSequences :: Int -> Int -> Integer
countSequences 0 m = 0
countSequences m 0 = 0
countSequences n m =
  if   n <=  m + 1
  then truncate $
         cee (fromIntegral (m+1)) (fromIntegral n) 0 +
         fromIntegral (countSequences (n-1) m)
  else countSequences (n-1) m

edited Dec 20, 2011 at 22:13

answered Dec 20, 2011 at 22:07

Stefan Holdermans

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