# 07.26.07

## Exercise 5.2

map map [x] = [map x] |

Therefore:

x :: a -> b |

because `x` is the first argument to `map`. Consider the type of map:

map :: (a -> b) -> [a] -> [b] |

This implies:

map x :: [a] -> [b] [map x] :: [[a] -> [b]] |

And hence:

map map :: [a -> b] -> [[a] -> [b]] |

–

foldl foldl x [y] = ? |

This is a type error. See technical error 10.

–

map foldl [x] = [foldl x] |

And as before:

foldl :: (a -> b -> a) -> a -> [b] -> a x :: a -> b -> a foldl x :: a -> [b] -> a [foldl x] :: [a -> [b] -> a] map foldl :: [a -> b -> a] -> [a -> [b] -> a] |

Mike said,

March 29, 2009 at 10:29 pm

I do not understand the first one, map map. The first argument to map is a function (a->b). I don’t see how “map” can serve as the first argument. Hope you can help. I don’t follow the leap from the second last line to the conclusion.

admin said,

March 29, 2009 at 10:55 pm

Using different type variables to make it clearer, the type (a -> b) can be unified with the type (c -> d) -> [c] -> [d] by considering a = (c -> d) and b = [c] -> [d]. So it is perfectly OK for map to be the first argument to map.