In Python and Ruby (and others as well, I'm sure). you can prefix an enumerable with * ("splat") to use it as an argument list. For instance, in Python:
>>> def foo(a,b): return a + b
>>> foo(1,2)
3
>>> tup = (1,2)
>>> foo(*tup)
3
Is there something similar in Haskell? I assume it wouldn't work on lists due to their arbitrary length, but I feel that with tuples it ought to work. Here's an example of what I'd like:
ghci> let f a b = a + b
ghci> :t f
f :: Num a => a -> a -> a
ghci> f 1 2
3
ghci> let tuple = (1,2)
I'm looking for an operator (or function) that allows me to do:
ghci> f `op` tuple
3
I have seen (<*>) being called "splat", but it doesn't seem to be referring to the same thing as splat in other languages. I tried it anyway:
ghci> import Control.Applicative
ghci> f <*> tuple
<interactive>:1:7:
Couldn't match expected type `b0 -> b0'
with actual type `(Integer, Integer)'
In the second argument of `(<*>)', namely `tuple'
In the expression: f <*> tuple
In an equation for `it': it = f <*> tuple
Yes, you can apply functions to tuples, using the tuple package. Check out, in particular, the uncurryN function, which handles up to 32-tuples:
Prelude Data.Tuple.Curry> (+) `uncurryN` (1, 2)
3
The uncurry function turns a function on two arguments into a function on a tuple. Lists would not work in general because of their requirement for homogeneity.
No, Haskell's type system doesn't like that. Check this similar question for more details:
How do I define Lisp’s apply in Haskell?
BTW, the splat operator you talk about is also known as the apply function, commonly found on dynamical functional languages (like LISP and Javascript).
Related
Since I had trouble googling this question I thought I'd post it here.
I'm just interested in the logic behind it or wether it's just the creators' preference to use ++ instead. I mean, using a typeclass for strings that concatenates two strings (or rather lists) with + does not seem too hard to imagine.
Edit: I should add, that in Haskell one has to suspect reasons behind it, because + and ++ are functions defined in typeclasses, whereas in java the usage of + for string concatenation is just part of the language's syntax and therefor subject only to the creators preference/opinion. (The answers so far suggest that I was right about my suspicion.)
Also haskell comes from a mathematical background and is deeply influenced by mathematical syntax, so there might be deeper reasons than just preference/opinion.
typeclass for strings that concatenates two strings
Such a typeclass exists, although the operator isn't +, but <>:
Prelude> :m +Data.Monoid
Prelude Data.Monoid> "foo" <> "bar"
"foobar"
While ++ concatenates lists, the <> operator is more general, since it combines any two values of a given Monoid instance.
As other people have pointed out, + is reserved for Num instances. Why isn't the Monoid binary operator called +, then? Because addition is only one of infinitely many monoids; multiplication is another:
Prelude Data.Monoid> Sum 2 <> Sum 3
Sum {getSum = 5}
Prelude Data.Monoid> Product 2 <> Product 3
Product {getProduct = 6}
Choosing something like <> as 'the' monoidal operator is preferred exactly because it carries little semantic baggage.
Long story short, it would cause type troubles.
(+) is part of the Num typeclass:
class Num a where
(+), (-), (*) :: a -> a -> a
negate :: a -> a
abs :: a -> a
signum :: a -> a
fromInteger :: Integer -> a
x - y = x + negate y
negate x = 0 - x
And (++) :: [a] -> [a] -> [a].
It's easy to see the first problem: if we wanted (+) to work on list, we would have to implement (*), negate, abs, signum, and fromInteger for lists as well. Which is spurious.
If we decided to seperate (+) from the typeclass, and make a new typeclass, maybe called Plussable for (+), there would be too many typeclasses to keep track of, and simple expressions like 1 + 2*(2-1) would no longer be of type Num a => a, it would be of type (Plussable a, Timesable a, Minusable a) => a, and so on for each operation. It would be far too complicated.
In Python and Ruby (and others as well, I'm sure). you can prefix an enumerable with * ("splat") to use it as an argument list. For instance, in Python:
>>> def foo(a,b): return a + b
>>> foo(1,2)
3
>>> tup = (1,2)
>>> foo(*tup)
3
Is there something similar in Haskell? I assume it wouldn't work on lists due to their arbitrary length, but I feel that with tuples it ought to work. Here's an example of what I'd like:
ghci> let f a b = a + b
ghci> :t f
f :: Num a => a -> a -> a
ghci> f 1 2
3
ghci> let tuple = (1,2)
I'm looking for an operator (or function) that allows me to do:
ghci> f `op` tuple
3
I have seen (<*>) being called "splat", but it doesn't seem to be referring to the same thing as splat in other languages. I tried it anyway:
ghci> import Control.Applicative
ghci> f <*> tuple
<interactive>:1:7:
Couldn't match expected type `b0 -> b0'
with actual type `(Integer, Integer)'
In the second argument of `(<*>)', namely `tuple'
In the expression: f <*> tuple
In an equation for `it': it = f <*> tuple
Yes, you can apply functions to tuples, using the tuple package. Check out, in particular, the uncurryN function, which handles up to 32-tuples:
Prelude Data.Tuple.Curry> (+) `uncurryN` (1, 2)
3
The uncurry function turns a function on two arguments into a function on a tuple. Lists would not work in general because of their requirement for homogeneity.
No, Haskell's type system doesn't like that. Check this similar question for more details:
How do I define Lisp’s apply in Haskell?
BTW, the splat operator you talk about is also known as the apply function, commonly found on dynamical functional languages (like LISP and Javascript).
So I'm playing around with the hasbolt module in GHCi and I had a curiosity about some desugaring. I've been connecting to a Neo4j database by creating a pipe as follows
ghci> pipe <- connect $ def {credentials}
and that works just fine. However, I'm wondering what the type of the (<-) operator is (GHCi won't tell me). Most desugaring explanations describe that
do x <- a
return x
desugars to
a >>= (\x -> return x)
but what about just the line x <- a?
It doesn't help me to add in the return because I want pipe :: Pipe not pipe :: Control.Monad.IO.Class.MonadIO m => m Pipe, but (>>=) :: Monad m => m a -> (a -> m b) -> m b so trying to desugar using bind and return/pure doesn't work without it.
Ideally it seems like it'd be best to just make a Comonad instance to enable using extract :: Monad m => m a -> a as pipe = extract $ connect $ def {creds} but it bugs me that I don't understand (<-).
Another oddity is that, treating (<-) as haskell function, it's first argument is an out-of-scope variable, but that wouldn't mean that
(<-) :: a -> m b -> b
because not just anything can be used as a free variable. For instance, you couldn't bind the pipe to a Num type or a Bool. The variable has to be a "String"ish thing, except it never is actually a String; and you definitely can't try actually binding to a String. So it seems as if it isn't a haskell function in the usual sense (unless there is a class of functions that take values from the free variable namespace... unlikely). So what is (<-) exactly? Can it be replaced entirely by using extract? Is that the best way to desugar/circumvent it?
I'm wondering what the type of the (<-) operator is ...
<- doesn't have a type, it's part of the syntax of do notation, which as you know is converted to sequences of >>= and return during a process called desugaring.
but what about just the line x <- a ...?
That's a syntax error in normal haskell code and the compiler would complain. The reason the line:
ghci> pipe <- connect $ def {credentials}
works in ghci is that the repl is a sort of do block; you can think of each entry as a line in your main function (it's a bit more hairy than that, but that's a good approximation). That's why you need (until recently) to say let foo = bar in ghci to declare a binding as well.
Ideally it seems like it'd be best to just make a Comonad instance to enable using extract :: Monad m => m a -> a as pipe = extract $ connect $ def {creds} but it bugs me that I don't understand (<-).
Comonad has nothing to do with Monads. In fact, most Monads don't have any valid Comonad instance. Consider the [] Monad:
instance Monad [a] where
return x = [x]
xs >>= f = concat (map f xs)
If we try to write a Comonad instance, we can't define extract :: m a -> a
instance Comonad [a] where
extract (x:_) = x
extract [] = ???
This tells us something interesting about Monads, namely that we can't write a general function with the type Monad m => m a -> a. In other words, we can't "extract" a value from a Monad without additional knowledge about it.
So how does the do-notation syntax do {x <- [1,2,3]; return [x,x]} work?
Since <- is actually just syntax sugar, just like how [1,2,3] actually means 1 : 2 : 3 : [], the above expression actually means [1,2,3] >>= (\x -> return [x,x]), which in turn evaluates to concat (map (\x -> [[x,x]]) [1,2,3])), which comes out to [1,1,2,2,3,3].
Notice how the arrow transformed into a >>= and a lambda. This uses only built-in (in the typeclass) Monad functions, so it works for any Monad in general.
We can pretend to extract a value by using (>>=) :: Monad m => m a -> (a -> m b) -> m b and working with the "extracted" a inside the function we provide, like in the lambda in the list example above. However, it is impossible to actually get a value out of a Monad in a generic way, which is why the return type of >>= is m b (in the Monad)
So what is (<-) exactly? Can it be replaced entirely by using extract? Is that the best way to desugar/circumvent it?
Note that the do-block <- and extract mean very different things even for types that have both Monad and Comonad instances. For instance, consider non-empty lists. They have instances of both Monad (which is very much like the usual one for lists) and Comonad (with extend/=>> applying a function to all suffixes of the list). If we write a do-block such as...
import qualified Data.List.NonEmpty as N
import Data.List.NonEmpty (NonEmpty(..))
import Data.Function ((&))
alternating :: NonEmpty Integer
alternating = do
x <- N.fromList [1..6]
-x :| [x]
... the x in x <- N.fromList [1..6] stands for the elements of the non-empty list; however, this x must be used to build a new list (or, more generally, to set up a new monadic computation). That, as others have explained, reflects how do-notation is desugared. It becomes easier to see if we make the desugared code look like the original one:
alternating :: NonEmpty Integer
alternating =
N.fromList [1..6] >>= \x ->
-x :| [x]
GHCi> alternating
-1 :| [1,-2,2,-3,3,-4,4,-5,5,-6,6]
The lines below x <- N.fromList [1..6] in the do-block amount to the body of a lambda. x <- in isolation is therefore akin to a lambda without body, which is not a meaningful thing.
Another important thing to note is that x in the do-block above does not correspond to any one single Integer, but rather to all Integers in the list. That already gives away that <- does not correspond to an extraction function. (With other monads, the x might even correspond to no values at all, as in x <- Nothing or x <- []. See also Lazersmoke's answer.)
On the other hand, extract does extract a single value, with no ifs or buts...
GHCi> extract (N.fromList [1..6])
1
... however, it is really a single value: the tail of the list is discarded. If we want to use the suffixes of the list, we need extend/(=>>)...
GHCi> N.fromList [1..6] =>> product =>> sum
1956 :| [1236,516,156,36,6]
If we had a co-do-notation for comonads (cf. this package and the links therein), the example above might get rewritten as something in the vein of:
-- codo introduces a function: x & f = f x
N.fromList [1..6] & codo xs -> do
ys <- product xs
sum ys
The statements would correspond to plain values; the bound variables (xs and ys), to comonadic values (in this case, to list suffixes). That is exactly the opposite of what we have with monadic do-blocks. All in all, as far as your question is concerned, switching to comonads just swaps which things we can't refer to outside of the context of a computation.
Suppose I have a value of type Monad m => (m a, m a), and I want to "sequence" the pair to create a value of type Monad m => m (a, a) that combines the monadic context of the two values in the same way the "sequence" function does. Is there some standard function or standard way of doing this? And does this operation even make sense?
ghci> import Control.Lens
ghci> sequenceOf both (getLine, getLine)
Apples
Bananas
("Apples","Bananas")
There wouldn't be a single function for all the different tuple types, since it wouldn't have a single type.
You could define a family of functions like:
ts0 = return
ts2 = uncurry $ liftM2 (,)
ts3 = uncurr3 $ liftM3 (,,)
{- ... -}
uncurr3 f (x, y, z) = f x y z
Of course, sequence in general is better applied to applicatives instead of monads, which is why it is part of the Traversable typeclass. It would be possible to make homogenous tuples [(a,a,a) but not (a,b,a)] an instance of MonoTraversable, I believe.
You should also see another answer which indicates there is already a library containing this family of functions.
The tuple package has Data.Tuple.Sequence.sequenceT which is overloaded for up to 32-tuples.
I have to dessignate types of 2 functions(without using compiler :t) i just dont know how soudl i read these functions to make correct steps.
f x = map -1 x
f x = map (-1) x
Well i'm a bit confuse how it will be parsed
Function application, or "the empty space operator" has higher precedence than any operator symbol, so the first line parses as f x = map - (1 x), which will most likely1 be a type error.
The other example is parenthesized the way it looks, but note that (-1) desugars as negate 1. This is an exception from the normal rule, where operator sections like (+1) desugar as (\x -> x + 1), so this will also likely1 be a type error since map expects a function, not a number, as its first argument.
1 I say likely because it is technically possible to provide Num instances for functions which may allow this to type check.
For questions like this, the definitive answer is to check the Haskell Report. The relevant syntax hasn't changed from Haskell 98.
In particular, check the section on "Expressions". That should explain how expressions are parsed, operator precedence, and the like.
These functions do not have types, because they do not type check (you will get ridiculous type class constraints). To figure out why, you need to know that (-1) has type Num n => n, and you need to read up on how a - is interpreted with or without parens before it.
The following function is the "correct" version of your function:
f x = map (subtract 1) x
You should be able to figure out the type of this function, if I say that:
subtract 1 :: Num n => n -> n
map :: (a -> b) -> [a] -> [b]
well i did it by my self :P
(map) - (1 x)
(-)::Num a => a->a->->a
1::Num b=> b
x::e
map::(c->d)->[c]->[d]
map::a
a\(c->d)->[c]->[d]
(1 x)::a
1::e->a
f::(Num ((c->d)->[c]->[d]),Num (e->(c->d)->[c]->[d])) => e->(c->d)->[c]->[d]