[Haskell-cafe] What is a rigid type variable?
Xiao-Yong Jin
xj2106 at columbia.edu
Sun Jun 22 23:26:58 EDT 2008
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/*
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