Efficient way of printing property names in Haskell

Hello i am new to Haskell and i was wondering :

For every new defined type constructor even if it is something with 0 depth eg.:

data Letter =A | B | C deriving (Show)


do i have to provide

instance Show Type
Letter A = "A"
Letter B ="B"
Letter C ="C"


I understand that for nested types you have to provide Show implementation but for something simple is there no other way to just use something equivalent to reflection (C#) ? Just get the property name / constructor name and ToString-it ?

Show

ToString

Do haskell programmers provide instances for Show for any adhoc type they create?

Show

By nested types i mean having a simple ADT inside another ADT inside another etc..but all you want is to get their names:

e.g:

data Person = Male | Female
data IsMarried=Yes | No

data Worker=Worker{
person::Person,
status::IsMarried
}


For this kind of nesting do i really have to define instance of Show for all these type constructors even though all i want is their name be "stringified"?

instance

Show

You have to instruct GHC how to render your type, yes. If you use deriving Show you do not need to provide an instance of Show.
– Robert K
Jun 29 at 9:29

deriving Show

Show

If you add deriving Show the compiler writes the instance for you. Unless you want it to be different from the compiler's default, there's no need to write it on your own.
– chi
Jun 29 at 11:32

deriving Show

2 Answers
2

Since @AJFarmar edit back his answer when i change it:

Do I have to provide [a show instance for every type?]

Yes. Either by deriving it with deriving Show, or by supplying a type instance

deriving Show

instance Show -type- where...

instance Show -type- where...

I understand that for nested types you have to provide Show implementation[...]

Yes you do, the compiler will not add any necessary instances for Show.

data Test = Test -- no instance for Show

test :: String
test = show $ (Just Test :: Maybe Test)


Will not compile with the error message:

main.hs:4:8: error:
• No instance for (Show Test) arising from a use of ‘show’
• In the expression: show $ (Just Test :: Maybe Test)
In an equation for ‘test’: test = show $ (Just Test :: Maybe Test)


Which is why you need to have a Show instance for a aswell. If one already exists, you do not have to supply a new one, however.

Show

a

Do I have to provide [a show instance for every type?]

No, because you have automatically derived Show:

Show

data Letter = A | B | C deriving (Show)
-- ^^^^^^^^^^^^^^^ Here


However, if you want a 'smarter' show, such as displaying a custom list as [1,2,3] rather than a mess of constructors, you're going to have to write it yourself.

show

[1,2,3]

You can do the same for a number of classes (Including Eq,Ord,Read,Enum, and Bounded) but most classes, including user-defined classes, must be implemented manually without the use of certain language extensions.

Eq

Ord

Read

Enum

Bounded

I understand that for nested types you have to provide Show implementation[...]

You do not! For instance, I can write this:

data Maybe a = Just a | Nothing deriving (Show)


And the compiler will automatically add the necessary constraints, despite it being a 'nested' type.

Just get the property name / constructor name and ToString-it ?

There are no 'properties' in Haskell - don't think in terms of C# here. show is the equivalent of ToString. However, a form of type reflection is available in TypeReps, though I would advise not using this until you have a solid grasp on Haskell.

show

ToString

You can not magically derive Show for any polymorphic data type. The polymorphic types must also have instances of Show.
– Robert K
Jun 29 at 9:48

Show

Show

@RobertK Quote: "And the compiler will automatically add the necessary constraints." I'm fully aware the a in Maybe a must also be an instance of Show.
– AJFarmar
Jun 29 at 9:53

a

Maybe a

Show

The compiler will not automatically add an instance of Show for a. This is where i claim you are wrong.
– Robert K
Jun 29 at 11:08

Show

a

@RobertK I suspect that what you mean by nested types is "types contained in another type", while what the original question asker meant was "types that contain another type". Perhaps this explains the confusion in the conversation thus far.
– Daniel Wagner
Jun 29 at 13:47

I am sorry for late response.For nested types i was meaning having ADT inside ADT. Something like data Person=Male | Female and data IsMarried=Yes | No in data Worker=Worker{ type::Person , status::IsMarried} .I didn't know if you have to provide show for each level of ADT even though you do not want any specific behaviour you just want the name of the data constructor toString.
– Bercovici Adrian
Jun 29 at 19:06

data Person=Male | Female

data IsMarried=Yes | No

data Worker=Worker{ type::Person , status::IsMarried}

toString

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