On Sun, 7 Mar 2010, Edgar Z. Alvarenga wrote:
Hello,
why I can't define a recursive vector using Data.Vector, like in
the example:
import qualified Data.Vector as V
let fib = 0 `V.cons` (1 `V.cons` V.zipWith (+) fib (V.tail v))
Since I liked to have both element-wise lazy construction and
edgar:
Hello,
why I can't define a recursive vector using Data.Vector, like in
the example:
import qualified Data.Vector as V
let fib = 0 `V.cons` (1 `V.cons` V.zipWith (+) fib (V.tail v))
There's a typo:
fib = 0 `V.cons` (1 `V.cons` V.zipWith (+) fib (V.tail fib))
Which let's it
On Mar 7, 2010, at 12:56 PM, Don Stewart wrote:
In fact, infinite vectors make no sense, as far as I can tell -- these
are fundamentally bounded structures.
Fourier analysis? Functional analysis? Hamel bases in Real
analysis? There are lots of infinite dimensional vector spaces out
ajs:
On Mar 7, 2010, at 12:56 PM, Don Stewart wrote:
In fact, infinite vectors make no sense, as far as I can tell -- these
are fundamentally bounded structures.
Fourier analysis? Functional analysis? Hamel bases in Real analysis? There
are lots of infinite dimensional
On 08/03/2010, at 12:17, Alexander Solla wrote:
GHC even optimizes it to:
fib = fib
Sounds like an implementation bug, not an infinite dimensional vector space
bug. My guess is that strictness is getting in the way, and forcing what
would be a lazy call to fib in the corresponding
On Mar 7, 2010, at 5:22 PM, Don Stewart wrote:
Sorry for the overloading, I mean 'vector' in the sense of
Data.Vector.
Being strict in the length, its unclear to me that you can do much
with
infinite ones :-)
Yeah, fair enough. I studied mathematics, not Haskell's Data.*
hierarchy.
