The type (String -> Maybe (a, String)) should already be an instance of
arbitrary if a is.
One way of generating functions is to have some sort of hashing function (a
-> Int). Functions from a to b can be generating by using the hash value to
transform the random seed before the b is generated. Here is a rough outline
for such a system:
genB :: StdGen -> B
hashA :: A -> Int
transformGen :: Int -> StdGen -> StdGen
genFun :: StdGen -> (A -> B)
genFun g a = genB $ transformGen (hashA a) g
In QC you have this instance:
(CoArbitraryhttp://hackage.haskell.org/packages/archive/QuickCheck/2.1.0.1/doc/html/Test...
a, Arbitraryhttp://hackage.haskell.org/packages/archive/QuickCheck/2.1.0.1/doc/html/Test...
b)
=> Arbitraryhttp://hackage.haskell.org/packages/archive/QuickCheck/2.1.0.1/doc/html/Test...
(a
-> b)
And CoArbitrary is the class where you implement the trick described above.
Look at the function called variant as well.
/J
On 16 September 2010 02:59, Ivan Lazar Miljenovic wrote: On 16 September 2010 01:58, Thomas Davie Firstly, as far as i can tell, one cannot declare a type synonym to be
an instance of a type class, thus how would you make it an instance of
Arbitrary? The standard solution here is to create a newtype, and generate them
instead. I've also written and used dedicated generation functions (especially
for Strings, since I don't want _really_ arbitrary strings since some
of them might accidentally bring up formatting that my parser can't
cope with (or will parse as something else). Note that this is more
for generating sub-parts of other data types, rather than for types I
want to be able to run QC tests on. --
Ivan Lazar Miljenovic
Ivan.Miljenovic@gmail.com
IvanMiljenovic.wordpress.com
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