# Haskell/Solutions/Laziness

From Wikibooks

Exercises |
---|

- Why must we fully evaluate
`x` to normal form in`f x y = show x` ? - Which is the stricter function?
f x = length [head x] g x = length (tail x) |

- Because
`show x`

needs to evaluate`x`

to convert it to a`String`

. `g`

is stricter: it must walk through the whole list to discover its length, even if it does not need to examine the single values.

Instead, to evaluate `length [head x]`

, when `length`

evaluates the passed list it obtains `thunk_0:[]`

, where thunk_0 represents the result of `head x`

. Thus, you can verify that `f undefined => 1`

(actually, `f`

is equivalent to `const 1`

) while `g undefined`

fails. Also note that, if `ls`

is a valid `Int`

list (which does not contain `undefined`

among its elements), all following expressions fail to evaluate:

g (1:undefined) => length undefined => undefined

whereas the following evaluate successfully:

g (1:undefined:ls) => length (undefined:ls) => 1 + length ls g (undefined:ls) => length ls

The difference between `g (1:undefined)`

and `g (1:undefined:ls)`

may appear confusing, but is correct, because in the first case `undefined`

replaces a list (in particular, the tail of the cons cell passed to `g`

), while in the second case `undefined`

replaces a list value (which is not needed).