type Person = (NI, Age, Balance) type Bank = [Person] credit :: Bank -> [Person] credit [(a,b,c)] = [(a,b,c)] This code works when I type in: credit [(1,2,3)] but doesn't work when I type in: credit [(1,2,3),(4,5,6)] Any help? Thanks in advance. -- View this message in context: http://www.nabble.com/Newbie-list-question-tf3823011.html#a10823175 Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com.
junkywunky:
type Person = (NI, Age, Balance) type Bank = [Person]
credit :: Bank -> [Person] credit [(a,b,c)] = [(a,b,c)]
This code works when I type in:
credit [(1,2,3)]
but doesn't work when I type in:
credit [(1,2,3),(4,5,6)]
You're pattern matching in 'credit' on a list of a single element. Perhaps you mean to write: credit :: Bank -> [Person] credit x = x or perhaps return just the first element of the list: credit [] = [] credit (x:xs) = x You might want to start with one of the tutorials on Haskell programming listed on haskell.org. The wikibook is quite a good start. -- Don
junkywunky wrote:
type Person = (NI, Age, Balance) type Bank = [Person]
credit :: Bank -> [Person] credit [(a,b,c)] = [(a,b,c)]
This code works when I type in:
credit [(1,2,3)]
but doesn't work when I type in:
credit [(1,2,3),(4,5,6)]
Any help?
Thanks in advance.
The expression [(1,2,3),(4,5,6)] doesn't match the pattern [(a,b,c)]. Now, since Bank and [Person] are actually the exact same type and the credit function actually does nothing, you could simply write credit x = x (Or, for that matter, credit = id.) It would then work for both examples. I presume that the idea is that the credit function will eventually do something - in that case, it might be helpful to say exactly what you want it to actually do.
That's the thing. I want to return a list of people who are not overdrawn. Something like: type NI = Int type Age = Int type Balance = Int type Person = (NI, Age, Balance) type Bank = [Person] credit :: Bank -> [Person] credit [(a,b,c)] = [(a,b,c)] if c >= 0 then [(a,b,c)] else error "overdrawn customer" except this doesn't work with things like: credit [(1,2,3),(4,5,6)] or credit [(1,2,3),(4,5,6),(7,8,-9)] Andrew Coppin wrote:
junkywunky wrote:
type Person = (NI, Age, Balance) type Bank = [Person]
credit :: Bank -> [Person] credit [(a,b,c)] = [(a,b,c)]
This code works when I type in:
credit [(1,2,3)]
but doesn't work when I type in:
credit [(1,2,3),(4,5,6)]
Any help?
Thanks in advance.
The expression [(1,2,3),(4,5,6)] doesn't match the pattern [(a,b,c)].
Now, since Bank and [Person] are actually the exact same type and the credit function actually does nothing, you could simply write
credit x = x
(Or, for that matter, credit = id.) It would then work for both examples.
I presume that the idea is that the credit function will eventually do something - in that case, it might be helpful to say exactly what you want it to actually do.
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-- View this message in context: http://www.nabble.com/Newbie-list-question-tf3823011.html#a10823244 Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com.
junkywunky:
That's the thing. I want to return a list of people who are not overdrawn. Something like:
type NI = Int type Age = Int type Balance = Int type Person = (NI, Age, Balance) type Bank = [Person]
credit :: Bank -> [Person] credit [(a,b,c)] = [(a,b,c)] if c >= 0 then [(a,b,c)] else error "overdrawn customer"
except this doesn't work with things like:
Right, you mean to write a list filter. List comprehensions are useful for this: credit xs = [ p | p@(a,b,c) <- xs, c >= 0 ] or maybe: credit xs = filter ok xs where ok (a,b,c) = c >= 0 -- Don
type Person = (NI, Age, Balance) type Bank = [Person]
credit :: Bank -> [Person] credit [(a,b,c)] = [(a,b,c)] if c >= 0 then [(a,b,c)] else error "overdrawn customer"
except this doesn't work with things like:
credit [(1,2,3),(4,5,6)]
Hi, that's because Haskell syntax is made for brains with high modality. When you declare a type, writing a type signature in square brackets make it to be a list of arbitrary number of elements of the inner type; when you write a pattern, one with an element in square brackets matches a single-element list. What you want is to credit abcs = filter (\(a,b,c) -> c>=0) abcs And if you think it looks like a machine-level assembly language, then you are probably right. David
participants (4)
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Andrew Coppin -
David Tolpin -
dons＠cse.unsw.edu.au -
junkywunky