learning-haskell/27-nonstrictness/27-nonstrictness.md

27 Nonstrictness

27.5 Exercises: Evaluate

Excercise 1

const 1 undefined
(\a -> \b -> a) 1 undefined
(\b -> 1) undefined
1

Exercise 2

const undefined 1
(\a -> \b -> a) undefined 1
(\b -> undefined) 1
undefined

Exercise 3

flip const undefined 1
(\f -> \a -> \b -> f b a) const undefined 1
(\a -> \b -> const b a) undefined 1
(\b -> const b undefined) 1
const 1 undefined -- same as Exercise 1 now

Exercise 4 follows same principle as Exercise 3 but leads to Exercise 2.

Exercise 5 follows same pattern as before but where 1 is undefined.

Exercise 6

foldr const 'z' ['a'..'e']
const 'a' (foldr const 'z' ['b'..'e'])
'a'

Exercise 7

foldr (flip const) 'z' ['a'..'e']
(flip const) 'a' (foldr (flip const) 'z' ['b'..'e'])
const (foldr (flip const) 'z' ['b'..'e']) 'a'

foldr (flip const) 'z' ['b'..'e']
(flip const) 'b' (foldr (flip const) 'z' ['c'..'e'])
const (foldr (flip const) 'z' ['c'..'e']) 'b'

foldr (flip const) 'z' ['c'..'e']
(flip const) 'c' (foldr (flip const) 'z' ['d','e'])
const (foldr (flip const) 'z' ['d','e']) 'c'

foldr (flip const) 'z' ['d','e']
(flip const) 'd' (foldr (flip const) 'z' ['e'])
const (foldr (flip const) 'z' ['e']) 'd'

foldr (flip const) 'z' ['e']
(flip const) 'e' (foldr (flip const) 'z' [])
const (foldr (flip const) 'z' []) 'e'

foldr (flip const) 'z' []
'z'

27.14 Chapter Exercises

What will :sprint output

1. x = _
2. x = "1"
3. x = _
4. x = 1
5. x = _
6. x = _

Will printing this expression result in bottom

1. No
2. Yes
3. Yes
4. No
5. No
6. No
7. Yes

Make the expression bottom

There are different ways of doing this. This is one of them:

x = undefined
y = "blah"
main = do
    print (snd (x `seq` (x,y)))