[Haskell-cafe] Re: [Haskell] Strictness question
Ben.Lippmeier at anu.edu.au
Tue Jun 7 22:59:28 EDT 2005
>> To gloss over details: it'll reduce x far enough so it knows that
>> it's an Integer, but it won't nessesarally compute that integers
> No, Integers don't contain any lazy components.
> It statically knows that it's an integer.
I meant that it would reduce to the outermost constructor but
nessesarally evaluate the rest of the object.
Ok, I actually looked up the implementation of Integer in GHC.
> -- | Arbitrary-precision integers.
> data Integer
> = S# Int# -- small integers
> #ifndef ILX
> | J# Int# ByteArray# -- large integers
> | J# Void BigInteger -- .NET big ints
You were right and I was wrong, Integers contain no lazy components.
Perhaps that just highlights the folly of guessing how much actually
gets evaluated in a lazy language.. :)
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