To answer the question in the subject:
From "Simple unification-based type inference for GADTs", Peyton-Jones, et al. ICFP 2006. http://research.microsoft.com/users/simonpj/papers/gadt/
"Instead of "user-specified type", we use the briefer term rigid
type to describe a type that is completely specified, in some
direct fashion, by a programmer-supplied type annotation."
So a rigid type is any type specified by a programmer type signature.
All other types are "wobbly".
Does anyone know what is going to change about the terminology with
the new "boxy types" paper?
http://research.microsoft.com/users/simonpj/papers/boxy/
-- ryan
-- ryan
On Sun, Jun 22, 2008 at 8:26 PM, Xiao-Yong Jin
Hi all,
I'm writing a short function as follows, but I'm not able to find a suitable type signature for `go'. It uses Numeric.LinearAlgebra from hmatrix.
-- | Map each element in a vector to vectors and thus form a matrix -- | row by row mapVecToMat :: (Element a, Element b) => (a -> Vector b) -> Vector a -> Matrix b mapVecToMat f v = fromRows $ go (d - 1) [] where d = dim v go :: Element b => Int -> [Vector b] -> [Vector b] go 0 vs = f (v @> 0) : vs go !j !vs = go (j - 1) (f (v @> j) : vs)
If I give the type signature to go as this, I got the following error
Couldn't match expected type `b1' against inferred type `b' `b1' is a rigid type variable bound by the type signature for `go' at test.hs:36:20 `b' is a rigid type variable bound by the type signature for `mapVecToMat' at test.hs:31:35 Expected type: Vector b1 Inferred type: Vector b In the first argument of `(:)', namely `f (v @> 0)' In the expression: f (v @> 0) : vs
So what is this rigid type variable all about and what is correct type of the function `go'?
Thanks in advance, X-Y -- c/* __o/* <\ * (__ */\ < _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe