Re: [Haskell-cafe] Are casts required?

I always forget to reply all. Silly gmail. On Mon, Jun 6, 2011 at 2:07 AM, Ryan Ingram wrote:

Hi Pat. There aren't any casts in that code. There are type annotations, but this is different than the idea of a cast like in C.

For example ((3 :: Integer) :: Int) is a compile error.

What you are seeing is that 3 has the type (forall a. Num a => a); that is, the literal '3' gets converted by the compiler into

fromInteger (I# 3#)

where 3# represents the machine word '3' and I# is the internal constructor Word# -> Integer.

class Num a where ... fromInteger :: Integer -> a

So by 'casting', or rather, providing a type annotation, you are specifying what instance of Num gets the call to 'fromInteger'.

As to whether you *need* a type annotation: it depends. For example: foo () = sameId newId 3 the compiler will infer the type of 'foo' to be foo :: forall a. IDs a => () -> a

If you declare foo as a value, though, you run into the dreaded monomorphism restriction, and you might get a complaint from the compiler about ambiguity. foo2 = sameId newId 3

The monomorphism restriction forces values to be values; otherwise consider this

-- the usual 'expensive' computation fib :: Num a => a -> a fib 0 = 1 fib n = fib (n-1) + fib (n-2)

x = fib 100000

What's the type of x? Most generally, it's x :: Num a => a

But this means that x will be recalculated every time it's used; the value can't be saved since x doesn't represent a single value but rather a separate value for each instance of Num. You are allowed to manually specify this type, but without it, the compiler says 'You meant this to be a value!' and forces it to a particular type if it can, or complains about ambiguity if it can't. As to how it does so, look up the rules for defaulting and monomorphism in the Haskell report.

-- ryan

On Mon, Jun 6, 2011 at 12:45 AM, Patrick Browne wrote:

...

Are casts required to run the code below? If so why? Thanks, Pat

-- Idetifiers for objects class (Integral i) => IDs i where startId :: i newId :: i -> i newId i = succ i sameId, notSameId :: i -> i -> Bool -- Assertion is not easily expressible in Haskell -- notSameId i newId i = True sameId i j = i == j notSameId i j = not (sameId i j) startId = 1

instance IDs Integer where

-- are casts need here? sameId (newId startId::Integer) 3 sameId (3::Integer) (4::Integer) notSameId (3::Integer) (newId (3::Integer))

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