On 2015-09-10 09:23, Roelof Wobben wrote:

which I wants to test with this function :

prop_fourDifferent :: Integer -> Integer -> Integer -> Integer -> Bool prop_fourDifferent a b c d = fourDifferent a b c d == ( a == b ) && ( a == c ) && (a == d)

The problem is that (==) has a higher precedence (4) than (&&) (which has precedence 3). So your definition is equivalent to prop_fourDifferent a b c d = (fourDifferent a b c d == ( a == b )) && ( a == c ) && (a == d) You need some extra parentheses there, try prop_fourDifferent a b c d = fourDifferent a b c d == (( a == b ) && ( a == c ) && (a == d)) -- Frerich Raabe - raabe@froglogic.com www.froglogic.com - Multi-Platform GUI Testing

Hi Roelof,

False == True && False False

False == (True && False) True

:info (==) class Eq a where (==) :: a -> a -> Bool infix 4 ==

:info (&&) (&&) :: Bool -> Bool -> Bool infixr 3 &&

So (==) at level 4 binds tighter than (&&) at level 3. See: https://www.haskell.org/onlinereport/decls.html#fixity -- Kim-Ee On Thu, Sep 10, 2015 at 2:23 PM, Roelof Wobben wrote:

Hello,

I have this function :

fourDifferent:: Integer -> Integer -> Integer -> Integer -> Bool fourDifferent a b c d = ( a == b ) && ( a == c ) && (a == d)

which I wants to test with this function :

prop_fourDifferent :: Integer -> Integer -> Integer -> Integer -> Bool prop_fourDifferent a b c d = fourDifferent a b c d == ( a == b ) && ( a == c ) && (a == d)

so I do quickCheck propFourDifferent and see this outcome :

*Solution> quickCheck prop_fourDifferent *** Failed! Falsifiable (after 2 tests and 2 shrinks): 0 0 0 1

*Solution> fourDifferent 0 0 0 1 False *Solution> let a = 0 *Solution> let b = 0 *Solution> let c = 0 *Solution> let d = 1 *Solution> (a == b) && ( a == c) && ( a == d ) False *Solution>

So why is it failing. both gives false ,

Roelof

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