What does immutable variable in Haskell mean? - haskell

I am quite confused with the concept of immutable variables in Haskell. It seems like that we can't change the value of variables in Haskell. But when I tried following code in GHCI, it seemed like the value of variables did change:
Prelude> foo x=x+1
Prelude> a=1
Prelude> a
1
Prelude> foo a
2
Prelude> a=2
Prelude> a
2
Prelude> foo a
3
Does this conflict with the idea of immutable variables?
Many thanks!

Haskell doesn't allow you to modify existing variables. It does, however, allow you to re-use variable names, and that's all that's happening here. One way to see this is to ask GHCi, using the :i[nfo] directive, where the variable was declared:
Prelude> let a = 1
Prelude> :i a
a :: Num a => a -- Defined at <interactive>:2:5
Prelude> let a = 2
Prelude> :i a
a :: Num a => a -- Defined at <interactive>:4:5
These are actually two entire seperate, different variables, which just happen to be called the same name! If you just ask for a, the newer definition will be “preferred”, but the old one is still there – one way to see this, as remarked by chi in the comments, is to use a in a function:
Prelude> let a = 2
Prelude> :i a
a :: Num a => a -- Defined at <interactive>:4:5
Prelude> let f x = a + x
Prelude> let a = 3
Prelude> f (-2)
0
f never needs to care that you've defined a new variable that's also called a; from its perspective a was one immutable variable that always stays as it is.
It's worth talking a bit about why GHCi prefers the later definition. This does not otherwise happen in Haskell code; in particular if you try to compile the following module, it simply gives an error concerning duplicate definition:
a = 1
a = 2
main :: IO ()
main = print a
The reason that something like this is allowed in GHCi is that it works different from Haskell modules. The sequence of GHCi commands forms in fact a sequence of actions in the IO monad†; i.e. the program would have to be
main :: IO ()
main = do
let a = 1
let a = 2
print a
Now, if you've learned about monads you'll know that this is just syntactic sugar for
main =
let a = 1 in (let a = 2 in (print a))
and this is really the crucial bit for why you can re-use the name a: the second one, a = 2, lives in a narrower scope than the first. So it is more local, and local definitions have priority. Whether this is a good idea is a bit debatable; a good argument for it is that you can have a function like
greet :: String -> IO ()
greet name = putStrLn $ "Hello, "++name++"!"
and it won't stop working just because somebody defines elsewhere
name :: Car -> String
name car | rollsOverAtRightTurn car = "Reliant Robin"
| fuelConsumption car > 50*litrePer100km
= "Hummer"
| ... = ...
Besides, it's really quite useful that you can “redefine” variables while fooling around in GHCi, even though it's not such a good idea to redefine stuff in a proper program, which is supposed to show consistent behaviour.
†As dfeuer remarks, this is not the whole truth. You can do some things in GHCi that aren't allowed in an IO do-block, in particular you can define data types and classes. But any normal statement or variable definition act as it were in the IO monad.

(The other answer using GHCi is fine but to clarify it is not specific to GHCi or monads...)
As you can see from the fact that the following Haskell program
main =
let x = 1 in
let f y = x + y in
let x = 2 in
print (x * f 3)
prints 8 rather than 10, variables can only be "bound", not "mutated", in Haskell. In other words, the program above is equivalent (called α-equivalent, meaning a consistent change of the names of bound variables only; see https://wiki.haskell.org/Alpha_conversion for more details) to
main =
let x1 = 1 in
let f y = x1 + y in
let x2 = 2 in
print (x2 + f 3)
where it is clear that x1 and x2 are different variables.

Related

How does evaluation in Haskell work, for expressions with constraints

Suppose I write in GHCi:
GHCi> let x = 1 + 2 :: Integer
GHCi> seq x ()
GHCi> :sprint x
GHCi prints x = 3 as naturally expected.
However,
GHCi> let x = 1 + 2
GHCi> seq x ()
GHCi> :sprint x
yields x = _
The sole difference between the two expressions are their types (Integer vs Num a => a). My question is what exactly happens, and why is seemingly x not evaluated in the latter example.
The main issue is that
let x = 1 + 2
defines a polymorphic value of type forall a. Num a => a, and that is something which evaluates similarly to a function.
Each use of x can be made at a different type, e.g. x :: Int, x :: Integer, x :: Double and so on. These results are not "cached" in any way, but recomputed every time, as if x were a function which is called multiple times, so to speak.
Indeed, a common implementation of type classes implements such a polymorphic x as a function
x :: NumDict a -> a
where the NumDict a argument above is added by the compiler automatically, and carries information about a being a Num type, including how to perform addition, how to interpret integer literals inside a, and so on. This is called the "dictionary-passing" implementation.
So, using a polymorphic x multiple times indeed corresponds to invoking a function multiple times, causing recomputation. To avoid this, the (dreaded) Monomorphism Restriction was introduced in Haskell, forcing x to be monomorphic instead. The MR is not a perfect solution, and can create some surprising type errors in certain cases.
To alleviate this issue, the MR is disabled by default in GHCi, since in GHCi we don't care that much about performance -- usability is more important there. This however causes the recomputation to reappear, as you discovered.

How to implement static scope, dynamic scope, and lazy evaluation in ml and haskell?

I understand conceptually what all of these are, I'm just hoping for some code examples of how to implement them in ML and Haskell.
Haskell variables (top level definitions, variables in patterns, etc.) are all statically scoped. For example, the program:
y = "global value"
f = print y
g = let y = "local value" in f
main = g
will print "global value". Even though, in the definition of g, the function f is used after "redefining" y, this redefinition doesn't affect the definition of f which uses the statically (AKA lexically) scoped definition of y in force where f was defined.
If you want to "implement" dynamic scope, you have to be more specific about what you really mean. If you're wondering if you can write a function in plain Haskell, like:
addY :: Int -> Int
addY x = x + y
such that y might refer to a different variable from one call to the next, then the answer is no. In this definition, y always refers to the same variable (which, in Haskell, means the same, immutable value) which can be determined by static analysis of the program and cannot be dynamically redefined.
[[Edit: As #Jon Purdy points out, though, there's a Haskell extension that supports a form of dynamic scope such that the following prints various dynamically scoped local values with the same function.
{-# LANGUAGE ImplicitParams #-}
f :: (?y :: String) => IO ()
f = print ?y
g = let ?y = "g's local value" in f
h = let ?y = "h's local value" in f
main = do
g -- prints g's local value
h -- prints h's local value
let ?y = "main's local value" in f -- prints main's value
--end of edit--]]
For lazy evaluation, there are many examples, such as the following entered into an interactive GHCi session:
take 3 [1,2..] -- gives [1,2,3]
let x = (15^2, 6 `div` 0)
fst x -- gives 225
let y = snd x
y -- *** Exception: divide by zero
In the first line, if the evaluation was strict, the attempt to fully evaluate the infinite list [1,2..] (can also be written [1..] -- just counts up 1,2,3,.. forever) would go into an infinite loop, and the take function would never be called. In the second example, if evaluation was strict, the division by zero error would occur when x was defined, not only after we tried to print its second component.

Different numbers of arguments when pattern matching Maybe

I've run into a problem I don't really understand. I thought that I would be able to write code like this in Haskell:
foo :: Maybe Int -> Int
foo Nothing = 0
foo Just x = x
But when I try to compile it, I get the error:
Equations for ‘foo’ have different numbers of arguments
I can fix it by changing my code to the following:
foo :: Maybe Int -> Int
foo Nothing = 0
foo (Just x) = x
Which makes me think that GHC is interpreting Just as an argument to foo. But Haskell forbids using uppercase letters to start variable names, so I wouldn't think there should be any ambiguity here. What's going on?
You're correct, there's not ambiguity about whether or not Just is a constructor – but constructors can have no arguments! Haskell's pattern matching doesn't look up the names involved, it's strictly syntactic, and foo Just x = x is a perfectly well-formed function definition clause. It's ill-typed:
Prelude> let foo Just x = x
<interactive>:2:9:
Constructor ‘Just’ should have 1 argument, but has been given none
In the pattern: Just
In an equation for ‘foo’: foo Just x = x
but with different data types around, it'd be fine:
Prelude> data Justice = Just
Prelude> let foo Just x = x
Prelude> :t foo
foo :: Justice -> t -> t
Prelude> foo Just ()
()
Just could be a nullary constructor (as in the second example), and since function application is left-associative, the compiler parses Just and x as separate arguments and you get the "different numbers of arguments" error. (And as you can see above, if there weren't the Nothing case, you'd actually get the type error for code of that form.)
The idea is that pattern syntax should mirror application syntax. If I was calling foo I couldn't write foo Just x, because that means something else (and if foo had type (Int -> Maybe Int) -> Int -> Int then it would even work). Having patterns have different rules for where parentheses are needed than expressions would be very weird.
Writing compound patterns without parentheses and trusting the compiler to automatically group things also falls apart in more complex situations. What should this mean?
foo Just x : xs = ...

Defining variables inside a function Haskell

I am a huge newbie to Haskell, I actually just started 10 minutes ago. I am trying to figure out how to define a variable inside a function. Lets say I have the function
foo :: Int -> Int
foo a =
b = a * 2
b
-- Yes, I know, it doesn't do anything interesting
When I run it in GHCi I get a syntax error! How can you define a variable inside a function?
There are two ways to do this:
foo a = b where b = a * 2
foo a = let b = a * 2 in b
In most cases, the choice between them is an aesthetic rather than technical one. More precisely, where may only be attached to definitions, whereas let ... in ... may be used anywhere an expression is allowed. Both where and let introduce blocks, making multiple internal variables convenient in both cases.
Disregarding technical correctness, the answer is "sort of".
I think it's better to think of a variable as a function of zero arguments evaluating to a given value.
module Main where
import System.IO
foo :: Integer -> Integer
foo a =
b where
b = a * 2
main = do
putStrLn $ show $ foo 10

What type of scope does Haskell use?

I'm trying to figure out if Haskell uses dynamic or static scoping.
I realize that, for example, if you define:
let x = 10
then define the function
let square x = x*x
You have 2 different "x's", and does that mean it is dynamically scoped? If not, what scoping does it use, and why?
Also, can Haskell variables have aliases (a different name for the same memory location/value)?
Thanks.
Haskell use (broadly speaking) exactly the same lexical scoping as most other languages.
eg.
x = 10
Results in a value referenced through x in the global scope, whereas
square x = x * x
will result in x being lexically scoped to the function square. It may help if you think of the above form being a syntactic nicety for:
square = \ x -> x * x
As to your other question i'm not sure what you mean by aliasing
Answering only the second part of the question:
You can have several aliases for the same "memory location", but since they are all immutable, it does not matter most of the time.
Dumb example:
foo x y = x * y
bar z = foo z z
When within foo called from bar, both x and y are clearly the same value. But since you cannot modify either x or y, you will not even notice.
There are some things wrong in your statements...
There are no mutable variables in Haskell just definitions (or immutable variables)
A variable memory location is a concept that do not exist in Haskell
In your example, x is not 10 in the function is just a argument to square, that can take any value (you can specify the type later) in this case 10 but just in this case.
Here is an example of aliases provided by Curt Sampson:
import Data.IORef
main :: IO ()
main = do x <- newIORef 0 -- write 0 into x
readIORef x >>= print -- x contains 0
let y = x
readIORef y >>= print -- y contains 0
writeIORef x 42 -- write 42 into x
readIORef y >>= print -- y contains 42
As the first part of the question is already answered by others, here is the second part:
I assume by aliasing you mean one name for another. As haskell is a functional language, and functions behave as normal identifiers in any case, you can do that like this:
y = x
which would define an alias y for the function x. Note that everything is a function. Even if it looks like a "variable", it's just a nullary function taking no arguments. Aliases for types look like this:
type Function = Double -> Double
which would define an alias Function for the type Double -> Double
Haskell uses static nested scopes. What is a bit confusing compared with other languages that have static nested scopes is that the scope of a name is a block which includes tests preceding its definition. For example
evens = 0 : map (+1) odds
odds = map : (+1) evens
here the name 'odds' is in scope in the definition of 'evens', despite the surprising fact that 'odds' has not yet been defined. (The example defines two infinite lists of even and odd numbers.)
A dead language with a similar scoping rule was Modula-3. But Haskell is a bit trickier in that you can attempt to 'redefine' a variable within the same scope but instead you just introduce another recursion equation. This is a pitfall for people who learned ML or Scheme first:
let x = 2 * n
x = x + 1 -- watch out!
This is perfectly good ML or Scheme let*, but Haskel has scheme letrec semantics, without the restriction to lambda values. No wonder this is tricky stuff!
In your example, the global definition of x is shadowed by the local definition of x. In Haskell, a variable's scope is determined by a static reading of the source code - this is called lexical scope, but can get something similar to dynamic scoping with implicit parameters (but that can lead to some unexpected behavior (I've read; never tried 'em myself)).
To sum up the other answers concisely:
lexical scope
aliasing is as easy as x = 1; y = x but doesn't usually matter because things are immutable.
The let syntax you use in your example looks like it's at the interactive ghci> prompt. Everything in interactive mode occurs within the IO monad so things may appear more mutable there than normal.
Well, as I think people have said already, Haskell doesn't have any variables as found in most other languages, it only has expressions. In your example let x = 10 x is an expression that always evaluates to 10. You can't actually change the value of x later on, though you can use the scoping rules to hide it by defining x to be another expression.
Yes, Haskell has aliases. Try out this little program:
import Data.IORef
main :: IO ()
main = do x <- newIORef 0 -- write 0 into x
readIORef x >>= print -- x contains 0
let y = x
readIORef y >>= print -- y contains 0
writeIORef x 42 -- write 42 into x
readIORef y >>= print -- y contains 42

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