[Haskell-cafe] Newbie list question
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. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Newbie list question
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 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Newbie list question
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. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Newbie list question
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. > > ___ > Haskell-Cafe mailing list > Haskell-Cafe@haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe > > -- 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. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Newbie list question
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 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Newbie list question
Donald Bruce Stewart wrote: Don types faster than me. ;-) ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Newbie list question
> 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 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
