# Laziness is not always good

### From HaskellWiki

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− | Generally, since Haskell is a [[Non-strict_semantics|non-strict]] language, |
+ | Generally, since Haskell is a [[Non-strict_semantics|non-strict]] language, you should try to make a function [[maintaining laziness|least strict]]. |

− | you should try to make a function [[maintaining laziness|least strict]]. |
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This is in many cases the best semantics and the most efficient implementation. |
This is in many cases the best semantics and the most efficient implementation. |
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However, here is an important exception from the rule: |
However, here is an important exception from the rule: |
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forall a. mappend a mempty = a |
forall a. mappend a mempty = a |
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</haskell> |
</haskell> |
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− | You find that it is not <hask>mappend mempty undefined = undefined</hask>, |
+ | You find that it is not <hask>mappend mempty undefined = undefined</hask>, but <hask>mappend mempty undefined = mempty</hask>. |

− | but <hask>mappend mempty undefined = mempty</hask>. |
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Is this academic nitpicking or practically relevant? |
Is this academic nitpicking or practically relevant? |
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− | I think it is the latter one, because a <hask>Monoid</hask> instance implicitly promises |
+ | I think it is the latter one, because a <hask>Monoid</hask> instance implicitly promises that monoid laws can be applied in every case. |

− | that monoid laws can be applied in every case. |
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A programmer expects that every occurence of <hask>mappend mempty a</hask> can be safely replaced by <hask>a</hask>. |
A programmer expects that every occurence of <hask>mappend mempty a</hask> can be safely replaced by <hask>a</hask>. |
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You might even create an [[Playing by the rules|optimizer rule]] doing this. |
You might even create an [[Playing by the rules|optimizer rule]] doing this. |
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− | The above implementation of <hask>mappend</hask> however evaluates its operands lazily, |
+ | The above implementation of <hask>mappend</hask> however evaluates its operands lazily, and this gets lost when the optimization is applied. |

− | and this gets lost when the optimization is applied. |
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The solution of this issue is to define |
The solution of this issue is to define |
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* Haskell-Cafe on [http://www.haskell.org/pipermail/haskell-cafe/2009-January/054261.html Laws and partial values] |
* Haskell-Cafe on [http://www.haskell.org/pipermail/haskell-cafe/2009-January/054261.html Laws and partial values] |
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+ | * Haskell-Cafe on a space leak caused by [http://www.haskell.org/pipermail/haskell-cafe/2010-June/079444.html the garbage collector that did not recognize a selector-like function call] |
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* [[Maintaining laziness]] |
* [[Maintaining laziness]] |
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## Latest revision as of 16:21, 21 September 2011

Generally, since Haskell is a non-strict language, you should try to make a function least strict. This is in many cases the best semantics and the most efficient implementation. However, here is an important exception from the rule:

Consider theMonoid

()

mempty = () mappend _ _ = ()

These functions are least strict, but have a subtle problem: They do not generally satisfy the monoid laws.

Remind you:mempty

mappend

forall a. mappend mempty a = a forall a. mappend a mempty = a

mappend mempty undefined = undefined

mappend mempty undefined = mempty

Is this academic nitpicking or practically relevant?

I think it is the latter one, because aMonoid

mappend mempty a

a

You might even create an optimizer rule doing this.

The above implementation ofmappend

The solution of this issue is to define

mempty = () mappend () () = () force :: () -> () force _ = ()

and write

mappend (force a) (force b)

mappend a b

If you find that example too academic, you can choose any other data type with one constructor instead.

## [edit] 1 Exercise

Find out whether it would help to definemempty = undefined

## [edit] 2 See also

- Haskell-Cafe on Laws and partial values
- Haskell-Cafe on a space leak caused by the garbage collector that did not recognize a selector-like function call
- Maintaining laziness