# Num instance for functions

### From HaskellWiki

(Difference between revisions)

(more Humor than Proposal) |
(Formatting) |
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Line 2: | Line 2: | ||

to add functions nicely, say for | to add functions nicely, say for | ||

<haskell>f, g :: Num a => b -> a</haskell> | <haskell>f, g :: Num a => b -> a</haskell> | ||

− | + | you would define | |

<haskell>(f+g) x = f x + g x</haskell> | <haskell>(f+g) x = f x + g x</haskell> | ||

With an according definition of <hask>fromInteger</hask> | With an according definition of <hask>fromInteger</hask> | ||

<haskell>fromInteger = const</haskell> | <haskell>fromInteger = const</haskell> | ||

− | + | numeric literals would also denote constant functions. This allows | |

<haskell>f+2 == \x -> f x + 2</haskell>. | <haskell>f+2 == \x -> f x + 2</haskell>. | ||

Line 13: | Line 13: | ||

multiplication dot | multiplication dot | ||

<haskell>2(x+y) :: Integer</haskell> | <haskell>2(x+y) :: Integer</haskell> | ||

− | + | will now be parsed by a Haskell compiler to the most obvious meaning | |

<haskell>2 :: Integer</haskell> | <haskell>2 :: Integer</haskell> | ||

− | + | ! :-) | |

## Revision as of 13:10, 17 April 2007

Some people have argued, thatNum

(->)

to add functions nicely, say for

f, g :: Num a => b -> a

you would define

(f+g) x = f x + g x

fromInteger

fromInteger = const

numeric literals would also denote constant functions. This allows

f+2 == \x -> f x + 2

Even nicer, the mathematically established notation of omitting the multiplication dot

2(x+y) :: Integer

will now be parsed by a Haskell compiler to the most obvious meaning

2 :: Integer

! :-)