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103 lines
1.7 KiB
103 lines
1.7 KiB
7 years ago
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# 27 Nonstrictness
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## 27.5 Exercises: Evaluate
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Excercise 1
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```haskell
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const 1 undefined
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(\a -> \b -> a) 1 undefined
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(\b -> 1) undefined
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1
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```
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Exercise 2
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```haskell
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const undefined 1
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(\a -> \b -> a) undefined 1
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(\b -> undefined) 1
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undefined
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```
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Exercise 3
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```haskell
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flip const undefined 1
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(\f -> \a -> \b -> f b a) const undefined 1
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(\a -> \b -> const b a) undefined 1
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(\b -> const b undefined) 1
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const 1 undefined -- same as Exercise 1 now
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```
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Exercise 4 follows same principle as Exercise 3 but leads to Exercise 2.
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Exercise 5 follows same pattern as before but where 1 is undefined.
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Exercise 6
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```haskell
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foldr const 'z' ['a'..'e']
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const 'a' (foldr const 'z' ['b'..'e'])
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'a'
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```
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Exercise 7
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```haskell
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foldr (flip const) 'z' ['a'..'e']
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(flip const) 'a' (foldr (flip const) 'z' ['b'..'e'])
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const (foldr (flip const) 'z' ['b'..'e']) 'a'
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foldr (flip const) 'z' ['b'..'e']
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(flip const) 'b' (foldr (flip const) 'z' ['c'..'e'])
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const (foldr (flip const) 'z' ['c'..'e']) 'b'
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foldr (flip const) 'z' ['c'..'e']
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(flip const) 'c' (foldr (flip const) 'z' ['d','e'])
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const (foldr (flip const) 'z' ['d','e']) 'c'
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foldr (flip const) 'z' ['d','e']
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(flip const) 'd' (foldr (flip const) 'z' ['e'])
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const (foldr (flip const) 'z' ['e']) 'd'
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foldr (flip const) 'z' ['e']
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(flip const) 'e' (foldr (flip const) 'z' [])
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const (foldr (flip const) 'z' []) 'e'
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foldr (flip const) 'z' []
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'z'
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```
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## 27.14 Chapter Exercises
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### What will :sprint output
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1. `x = _`
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2. `x = "1"`
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3. `x = _`
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4. `x = 1`
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5. `x = _`
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6. `x = _`
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### Will printing this expression result in bottom
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1. No
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2. Yes
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3. Yes
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4. No
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5. No
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6. No
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7. Yes
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### Make the expression bottom
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There are different ways of doing this. This is one of them:
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```haskell
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x = undefined
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y = "blah"
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main = do
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print (snd (x `seq` (x,y)))
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```
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