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Message |
< Erlang ~ Simple exercises |
| pellee |
 Posted: Mon Sep 24, 2007 10:45 am |
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Joined: 24 Sep 2007
Posts: 1
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| thanks michal! |
Last edited by pellee on Mon Sep 24, 2007 11:48 am; edited 1 time in total |
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| michal |
| Posted: Mon Sep 24, 2007 11:41 am |
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User
Joined: 20 Jul 2006
Posts: 44
Location: London
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In below examples H is for header, T for tail and A for accumulator.
Solution to B could look like this:
Code:
reverse(List) ->
reverse(List, []).
reverse([H|T], A) -> reverse(T, [H|A]);
reverse([], A) -> A.
Solution to A could look like this:
Code:
filter(List, Integer) ->
filter(List, Integer, []).
filter([H|T], I, A) when H =< I -> filter(T, I, [H|A]);
filter([_|T], I, A) -> filter(T, I, A);
filter([], _, A) -> lists:reverse(A).
Filtering could be also solved using list comprehension:
Code:
filter2(List, Integer) ->
[Element || Element <- List,
Element =< Integer].
Michal |
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| francesco |
| Posted: Wed Sep 26, 2007 6:23 am |
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Joined: 07 Jul 2006
Posts: 249
Location: London
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