Confused about typing in ghci

Patrick Surry Patrick.Surry at portraitsoftware.com
Wed Apr 2 16:26:04 EDT 2008


I'm new to Haskell and functional programming; have been exploring the
fixed-point combinator based on the exercise in the HSOE book.  Having
an odd issue with typing in ghci which I don't understand - maybe it's
to do with section 3.4.5. "Type defaulting in GHCi" but I don't really
grok that yet...

 

(BTW, an unrelated question:  is there a good reference where I can
understand the precendence/associativity rules for Haskell?  I
continually find myself needing lots of parens before expressions behave
as I expect, and get very confused trying to use the $ operator -
usually resorting to lots of parens again :-)

 

My typing question might boil down to this simple example, though my
original example follows:

 

-- Why don't these (particularly g and g') all have the same type?

 

Prelude> :t (\x -> x+1)

(\x -> x+1) :: (Num a) => a -> a

Prelude> let g = (\x -> x+1)

Prelude> :t g

g :: Integer -> Integer

Prelude> let g' x = x + 1

Prelude> :t g'

g' :: (Num a) => a -> a

 

-- Here's my original fixed-point combinator example:

 

Prelude> let fix f = f (fix f)

 

-- here's a silly (but working) implementation of length using fix:

 

Prelude>  fix (\g xs -> if xs == [] then 0 else (1 + g (tail xs)))
[1,2,3,4]

4

 

-- so I examine the types of the parts, which seems fine:

 

Prelude> :t fix (\g xs -> if xs == [] then 0 else (1 + g (tail xs)))

...  :: (Num t, Eq a) => [a] -> t

 

Prelude> :t (\g xs -> if xs == [] then 0 else (1 + g (tail xs)))

... :: (Num t, Eq a) => ([a] -> t) -> [a] -> t

 

-- but now I try to bind the anonymous function to a name

-- this seems to get the types wrong and no longer works as I expect:

 

Prelude> let lenstep = (\g xs -> if xs == [] then 0 else (1 + g (tail
xs)))

Prelude> :t lenstep

lenstep :: ([()] -> Integer) -> [()] -> Integer

Prelude> :t fix lenstep

fix lenstep :: [()] -> Integer

Prelude> let len' = fix (\g xs -> if xs == [] then 0 else (1 + g (tail
xs)))

Prelude> :t len'

len' :: [()] -> Integer

Prelude>

Prelude> len' [1,2,3,4]

<interactive>:1:12:

    No instance for (Num ())

      arising from the literal `4' at <interactive>:1:12

    Possible fix: add an instance declaration for (Num ())

    In the expression: 4

    In the first argument of `len'', namely `[1, 2, 3, 4]'

    In the expression: len' [1, 2, 3, 4]

 

-- maybe this is just me not understanding name binding properly; it
seems to work if I do it this way:

 

Prelude> let lenstep' g xs = (if xs == [] then 0 else (1 + g (tail xs)))

Prelude> :t lenstep'

lenstep' :: (Eq a, Num t) => ([a] -> t) -> [a] -> t

Prelude> :t fix lenstep'

fix lenstep' :: (Eq a, Num t) => [a] -> t

Prelude> fix lenstep' [1,2,3,4]

4

 

-- but what's the difference?

 

Cheers,

Patrick

 

 

Patrick.Surry at portraitsoftware.com
<mailto:Patrick.Surry at portraitsoftware.com> , VP Technology 

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