Haskell: Property Based Testing for Higher Order Function

Question

I have two properties that a function foo must satisfy:

prop_1 :: [Int] -> Bool
prop_1 xs = foo xs id == xs 

prop_2 :: [Int] -> (Int -> Int) -> (Int -> Int) -> Bool
prop_2 xs f g = foo (foo xs f) g == foo xs (g . f)

I am trying to check whether the above properties satisfy the following function using quickCheck:

foo :: [a] -> (a -> b) -> [b]
foo xs f = []

When I tried running quickCheck with prop_2 I get the following error:

quickCheck(prop_2)

<interactive>:18:1: error:
     No instance for (Show (Int -> Int))
        arising from a use of 'quickCheck'
        (maybe you haven't applied a function to enough arguments?)
     In the expression: quickCheck (prop_2)
      In an equation for 'it': it = quickCheck (prop_2)

I am not sure why I am getting this error and how I can resolve it. Any insights are appreciated.

The arguments of prop_2 must be Showable, since QuickCheck should be able to show with what parameters the prop_2 was disproven. — willeM_ Van Onsem
– willeM_ Van Onsem, Commented Jun 28, 2019 at 10:14

Alexey Romanov · Accepted Answer · 2019-06-28 10:47:54Z

2

You can use QuickCheck's support for generation of random shrinkable, showable functions by changing the property to

prop_2 :: [Int] -> Fun Int Int -> Fun Int Int -> Bool
prop_2 xs (Fn f) (Fn g) = foo (foo xs f) g == foo xs (g . f)

and then you'll see something more useful than <function> for counterexamples.

answered Jun 28, 2019 at 10:47

Alexey Romanov

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2 Comments

ceno980 Over a year ago

What is <function> in quickCheck? I tried running quickCheck with another property and I got the following result: Failed (after 2 tests and 1 shrink) <function> "" "".

Alexey Romanov Over a year ago

It isn't "in QuickCheck", it's how the instance in Text.Show.Functions works (hackage.haskell.org/package/base-4.12.0.0/docs/…)

willeM_ Van Onsem · Accepted Answer · 2019-06-28 10:20:12Z

1

As the documentation on QuickCheck says:

However, before we can test such a property, we must see to it that function values can be printed (in case a counter-example is found). That is, function types must be instances of class Show. To arrange this, you must import module ShowFunctions into every module containing higher-order properties of this kind. If a counter-example is found, function values will be displayed as "<function>"

So you can fix this by importing a module like:

import Text.Show.Functions

prop_1 :: [Int] -> Bool
prop_1 xs = foo xs id == xs 

prop_2 :: [Int] -> (Int -> Int) -> (Int -> Int) -> Bool
prop_2 xs f g = foo (foo xs f) g == foo xs (g . f)

answered Jun 28, 2019 at 10:20

willeM_ Van Onsem

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1 Comment

dfeuer Over a year ago

It's really much better to use the Fun type as Alexey Romanov explains than to use a bogus orphan instance.

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