在Haskell中编辑数据类型

Question

i need help with haskell i'm kinda confused.

I have custom types/Data like this:

type Name         = String 
type LastName     = String
type Mail         = String
type FullName     = (LastName, Name)

data Person       = Person Mail FullName deriving (Show, Read)
type Contact      = (FullName,Mail)
type MailAccount = (Person,[Contact])

Let's assume that mail account is stored in a data_base , what i want from now is update the list of contact and i don't know how to do it using this signature:

updateContact :: Mail -> LastName -> Name ->MailAccount -> IO MailAccount

I tried with this :

updateContact l n m macc = do
           x <- createPerson l n m
           return $ mailAccountUpdate macc x

I created these three function:

 --return my list of contacts
  contacts:: MailAccount->[Contact]
  contacts (_,_,con) = con

  createPerson l n m = do
       return (Person m (l,n))

  mailAccountUpdate acc x = do 
       return (x:contact acc)

1. My problem is that thes code won't work because return $ mailAccountUpdate macc x retruns a list and not an IO MailAccount.
2. I'm not skilled enough to play with the functors and fmap yet.
3. I want a way to update my list of contact in the mail account , meaning i want a way to access and edit this data and override it with the updated list.
4. I have to respect the signature , so i tried to play with the logic in the terminal, so i tried a few things so my question is :

Is there a way to edit the data directly like OOP ex : MailAccount.contact() ? if not how can i create a function that can do the work.

Is there a way to have for example two mailAccount with the same type and code it to do the equivalent of this in the ghci terminal :

mail1 = mail2

This will override the data in mail1 with the data from mail2. but i dont know how to code it in haskell with data type.

Thank you everyone in advance for helping me.

Answer 1

There is a neater version of this but I'm doing like this so it can be more understanding:

type Name     = String
type LastName = String
type Mail     = String
type Id       = Int
type FullName = (LastName, Name)
type Contact  = (Id, FullName)

data Person =
  Person Mail FullName [Contact]
  deriving (Show, Read)

updatePersonContact :: Person -> FullName -> Id -> Person
updatePersonContact (Person m fn contacts) newfullName id =
    Person m fn (updateContact id newfullName contacts)

updateContact :: Id -> FullName -> [Contact] -> [Contact]
updateContact id newfullName contacts =
  map (\(i, fn) ->
         if i == id
         then (i, newfullName)
         else (i, fn)) contacts

person1 :: Person
person1 = Person "email@me.com"("last","first")
            [(1,("last1","first1")), (2,("last2","first2"))]

then using it you would:

> updatePersonContact person1 ("NEW","NEWW") 2
-- Person "email@me.com" ("last","first") [(1,("last1","first1")),(2,("NEW","NEWW"))]

We've update Person to have a list of [Contact] so now contacts are attached to a person. Each Contact now has an Id that we can use for getting to the right contact.

You can deconstruct your types in the input like so updatePersonContact (Person m fn contacts) so now we can use all the bits we need to reconstruct our person. With m fn we give them straight back as those don't need changing. But we're interested in contacts so we pass it to this function updateContact id newfullName contacts .

In updateContact as inputs we have an Id , FullName and list of [Contact] . the way to loop over a list is to use map so here we are using map to go through our list of contacts one by one. map () contacts

the () needs to be a function from a -> b and our a when it loops the first time is (1,("last1","first1") so as you can see it has 2 values which we deal with by having \\(i, fn) -> . So i == 1 and fn == ("last1","first1") . These would update on every iteration of the list like any loop.

The next part we check if i == id and if it does then that's the contact we want to update so we return (i, newfullName) which is our new full name originally passed into updatePersonContact . If i doesn't match id then let's just leave those values alone and return them as they came.

map is a functor so as you might see it's not that bad you might need to do a bit more reading on it to get a better understanding. :)

In Haskell lists are immutable which means every time you are given a list you must return a brand new list.

Answer 2

Check you contacts function. It should be

contacts (_,x) = x 

MailAccount is a tuple, with first element as person and second as contact. So it cannot be ( , ,x) but should be (_,x).

The below updatecontact definition is working fine.

updateContact l n m macc = let myP = Person l (n,m) 
                               newMacc = (myP,((n,m),l):contacts macc) 
                           in return newMacc

I agree with others that there is no need for using IO in definition. The error would have been understood easily without IO.