3 ms·
Answering to your edit: it is alternative a), your second code snippet is not a valid haskell value. The valid Haskell version would be: purelyFunctional :
by fhars 12y ago
Answering to your edit: it is alternative a), your second code snippet is not a valid haskell value. The valid Haskell version would be:
purelyFunctional :: IO Boolean
purelyFunctional = do
value1 <- notPurelyFunctional 5.0
value2 <- notPurelyFunctional 5.0
value1 == value2
Which, incidentally, is a pure value.
- dmytrish 12y agoThe valid Haskell version would be: ... return (value1 == value2)