I'm building comfort going through some Haskell toy problems and I've written the following speck of code
multipOf :: [a] -> (Int, a)
multipOf x = (length x, head x)
gmcompress x = (map multipOf).group $ x
which successfully preforms the following operation
gmcompress [1,1,1,1,2,2,2,3] = [(4,1),(3,2),(1,3)]
Now I want this function to instead of telling me that an element of the set had multiplicity 1, to just leave it alone. So to give the result [(4,1),(3,2),3] instead. It be great if there were a way to say (either during or after turning the list into one of pairs) for all elements of multiplicity 1, leave as just an element; else, pair. My initial, naive, thought was to do the following.
multipOf :: [a] -> (Int, a)
multipOf x = if length x = 1 then head x else (length x, head x)
gmcompress x = (map multipOf).group $ x
BUT this doesn't work. I think because the then and else clauses have different types, and unfortunately you can't piece-wise define the (co)domain of your functions. How might I go about getting past this issue?
BUT this doesn't work. I think because the then and else clauses have different types, and unfortunately you can't piece-wise define the (co)domain of your functions. How might I go about getting past this issue?
Your diagnosis is right; the then and else must have the same type. There's no "getting past this issue," strictly speaking. Whatever solution you adopt has to use same type in both branches of the conditional. One way would be to design a custom data type that encodes the possibilities that you want, and use that instead. Something like this would work:
-- | A 'Run' of #a# is either 'One' #a# or 'Many' of them (with the number
-- as an argument to the 'Many' constructor).
data Run a = One a | Many Int a
But to tell you the truth, I don't think this would really gain you anything. I'd stick to the (Int, a) encoding rather than going to this Run type.
I get an odd-looking warning when I use the (_,_) pattern in a list comprehension. My minimal working example is as follows.
theory Misc imports
Main
"~~/src/HOL/Library/Code_Target_Numeral"
begin
definition "xys = [(1::int,2::int),(2,3),(3,4)]"
value "[x+4. (x,_) ← xys]"
end
Everything seems to work fine, but I get the warning
The following clauses are redundant (covered by preceding clauses):
x ⇒ []
Should I be worried?
No need to be worried.
The syntax
[a+4. (a, b) ← xys]
is internally translated to
concat (map (λx. case x of (a, b) ⇒ [a + 4] | _ ⇒ []) xys)
(see the section "List comprehension" of HOL/List.thy).
What is happening here is that if your pattern (a, b) matches, it will be mapped to your expression a + 4 (the first half of the case expression). If it doesn't match, it will be dropped from the output list (the second half of the case expression). In your case, the pattern (a, b) will always match, meaning that the second half is redundant---hence the warning.
In the ideal case, the list comprehension implementation would be modified to not generate the second half of the case expression if the pattern entered by the user will always match. Until this is done, it is safe to just ignore the warning.
I'm trying to figure out the IO monad and the <- syntax which I often see in Haskell code. I've seen it used with multiple datatypes, including arrays and IO.
What is the equivalent operation in clojure if I were to specify one myself?
Do-notation is just sugar for the standard monad operations. For example, if you have something like this:
do
x <- someMonad
return (someFunction x)
That's equivalent to this:
someMonad >>= \x -> return (someFunction x)
So the equivalent Clojure using one of the many monad libraries might be something like this:
(m-bind some-monad (fn [x] (m-result (some-function x))))
Using algo.monads, we can define an IO monad easily (if unnecesarrily).
In Haskell, the IO monad is type IO a = World -> (a, World). It's handy to think of this as an action - something which takes the world, does something, and returns a value and the world.
Using a vector instead of a tuple, this means that, in Clojure, an IO action (a monadic value of the IO monad) looks something like this:
(fn [world]
; some stuff
[value world])
To do something interesting, we need a couple of actions: get-char and put-char.
get-char is an action which takes in the world, reads a char, and returns that char as its value alongside the world:
(defn read-char
[]
(-> *in* .read char))
(defn get-char
[world]
[(read-char) world])
put-char takes a character and creates an action which, given a world, prints the character and returns some (inconsequential) value:
(defn put-char
[c]
(fn [world]
(print c)
[nil world]))
Note that, to make an action happen, we have to supply a world. For instance, (put-char \a) will return an action; ((put-char \a) :world) will invoke that action, printing a and returning [nil :world].
Composing these actions is potentially a very messy process. If, for example, you wanted to get a character, then print it, you'd have to call get-char, unpack its character and world, create an action for that character with put-char, then pass the world to that action.
On the other hand, if we define a monad, we get domonad (the equivalent to Haskell's do) for free. This syntactic sugar alleviates the unpacking/packing boilerplate. We just need a few functions: m-result and m-bind (m-zero and m-plus are also handy, but not necessary).
m-result (return in Haskell) takes a value and wraps it up as an action:
(fn [v]
(fn [world]
[v world]))
m-bind (>>= in Haskell) takes an action and a function which takes a regular value to produce an action, "unwraps" the value by invoking the action, and applies the function to it. With the IO monad, that looks like this:
(fn [io f]
(fn [world]
(let [[v new-world] (io world)]
((f v) new-world))))
So, using algo.monads, we can define io-m as follows:
(defmonad io-m
[m-result (fn [v]
(fn [world]
[v world]))
m-bind (fn [io f]
(fn [world]
(let [[v new-world] (io world)]
((f v) new-world))))])
Now that we've got the primitive IO actions and a means of composing them, we can create more interesting ones. Note that Haskell's unpacking operator (<-) is implicit and the result is automatically wrapped with m-result so we don't use Haskell's return statement to terminate the expressions:
(declare get-rest-of-line)
(def get-line
(domonad io-m
[c get-char
line (if (= c \newline)
(m-result "")
(get-rest-of-line c))]
line))
(defn get-rest-of-line
[c]
(domonad io-m
[cs get-line]
(str c cs)))
(defn put-line
[s]
(if (seq s)
(domonad io-m
[_ (put-char (first s))
_ (put-line (subs s 1))]
_)
(put-char \newline)))
Finally, we can write a program in terms of these IO actions:
(def run-program
(domonad io-m
[line get-line
:let [reversed-line (->> line reverse (apply str))]
_ (put-line reversed-line)]
_))
(run-program :world)
I think Chuck has answered your main question, but in case you'd like to investigate the way the monad operations can be implemented in Clojure using algo.monads as an example, the following:
(domonad state-m
[_ (set-state "foo")
x (fetch-state)]
x)
is equivalent (well, almost, see below) to Haskell's
do
_ <- put "foo" -- see below for a comment on this
x <- get
return x
In algo.monads the <- disappears, because effectively it is implied on every line.
About the "almost" and the _ above: _ actually isn't magic in Clojure and it will be bound to the value returned by set-state, but it's idiomatic to use this symbol as the name of locals one doesn't care about. Of course in Haskell it would be more usual simply to write put "foo" instead of _ <- put "foo".
Here's a simple, barebones example of how the code that I'm trying to do would look in C++.
while (state == true) {
a = function1();
b = function2();
state = function3();
}
In the program I'm working on, I have some functions that I need to loop through until bool state equals false (or until one of the variables, let's say variable b, equals 0).
How would this code be done in Haskell? I've searched through here, Google, and even Bing and haven't been able to find any clear, straight forward explanations on how to do repetitive actions with functions.
Any help would be appreciated.
Taking Daniels comment into account, it could look something like this:
f = loop init_a init_b true
where
loop a b True = loop a' b' (fun3 a' b')
where
a' = fun1 ....
b' = fun2 .....
loop a b False = (a,b)
Well, here's a suggestion of how to map the concepts here:
A C++ loop is some form of list operation in Haskell.
One iteration of the loop = handling one element of the list.
Looping until a certain condition becomes true = base case of a function that recurses on a list.
But there is something that is critically different between imperative loops and functional list functions: loops describe how to iterate; higher-order list functions describe the structure of the computation. So for example, map f [a0, a1, ..., an] can be described by this diagram:
[a0, a1, ..., an]
| | |
f f f
| | |
v v v
[f a0, f a1, ..., f an]
Note that this describes how the result is related to the arguments f and [a0, a1, ..., an], not how the iteration is performed step by step.
Likewise, foldr f z [a0, a1, ..., an] corresponds to this:
f a0 (f a1 (... (f an z)))
filter doesn't quite lend itself to diagramming, but it's easy to state many rules that it satisfies:
length (filter pred xs) <= length xs
For every element x of filter pred xs, pred x is True.
If x is an element of filter pred xs, then x is an element of xs
If x is not an element of xs, then x is not an element of filter pred xs
If x appears before x' in filter pred xs, then x appears before x' in xs
If x appears before x' in xs, and both x and x' appear in filter pred xs, then x appears before x' in filter pred xs
In a classic imperative program, all three of these cases are written as loops, and the difference between them comes down to what the loop body does. Functional programming, on the contrary, insists that this sort of structural pattern does not belong in "loop bodies" (the functions f and pred in these examples); rather, these patterns are best abstracted out into higher-order functions like map, foldr and filter. Thus, every time you see one of these list functions you instantly know some important facts about how the arguments and the result are related, without having to read any code; whereas in a typical imperative program, you must read the bodies of loops to figure this stuff out.
So the real answer to your question is that it's impossible to offer an idiomatic translation of an imperative loop into functional terms without knowing what the loop body is doing—what are the preconditions supposed to be before the loop runs, and what the postconditions are supposed to be when the loop finishes. Because that loop body that you only described vaguely is going to determine what the structure of the computation is, and different such structures will call for different higher-order functions in Haskell.
First of all, let's think about a few things.
Does function1 have side effects?
Does function2 have side effects?
Does function3 have side effects?
The answer to all of these is a resoundingly obvious YES, because they take no inputs, and presumably there are circumstances which cause you to go around the while loop more than once (rather than def function3(): return false). Now let's remodel these functions with explicit state.
s = initialState
sentinel = true
while(sentinel):
a,b,s,sentinel = function1(a,b,s,sentinel)
a,b,s,sentinel = function2(a,b,s,sentinel)
a,b,s,sentinel = function3(a,b,s,sentinel)
return a,b,s
Well that's rather ugly. We know absolutely nothing about what inputs each function draws from, nor do we know anything about how these functions might affect the variables a, b, and sentinel, nor "any other state" which I have simply modeled as s.
So let's make a few assumptions. Firstly, I am going to assume that these functions do not directly depend on nor affect in any way the values of a, b, and sentinel. They might, however, change the "other state". So here's what we get:
s = initState
sentinel = true
while (sentinel):
a,s2 = function1(s)
b,s3 = function2(s2)
sentinel,s4 = function(s3)
s = s4
return a,b,s
Notice I've used temporary variables s2, s3, and s4 to indicate the changes that the "other state" goes through. Haskell time. We need a control function to behave like a while loop.
myWhile :: s -- an initial state
-> (s -> (Bool, a, s)) -- given a state, produces a sentinel, a current result, and the next state
-> (a, s) -- the result, plus resultant state
myWhile s f = case f s of
(False, a, s') -> (a, s')
(True, _, s') -> myWhile s' f
Now how would one use such a function? Well, given we have the functions:
function1 :: MyState -> (AType, MyState)
function2 :: MyState -> (BType, MyState)
function3 :: MyState -> (Bool, MyState)
We would construct the desired code as follows:
thatCodeBlockWeAreTryingToSimulate :: MyState -> ((AType, BType), MyState)
thatCodeBlockWeAreTryingToSimulate initState = myWhile initState f
where f :: MyState -> (Bool, (AType, BType), MyState)
f s = let (a, s2) = function1 s
(b, s3) = function2 s2
(sentinel, s4) = function3 s3
in (sentinel, (a, b), s4)
Notice how similar this is to the non-ugly python-like code given above.
You can verify that the code I have presented is well-typed by adding function1 = undefined etc for the three functions, as well as the following at the top of the file:
{-# LANGUAGE EmptyDataDecls #-}
data MyState
data AType
data BType
So the takeaway message is this: in Haskell, you must explicitly model the changes in state. You can use the "State Monad" to make things a little prettier, but you should first understand the idea of passing state around.
Lets take a look at your C++ loop:
while (state == true) {
a = function1();
b = function2();
state = function3();
}
Haskell is a pure functional language, so it won't fight us as much (and the resulting code will be more useful, both in itself and as an exercise to learn Haskell) if we try to do this without side effects, and without using monads to make it look like we're using side effects either.
Lets start with this structure
while (state == true) {
<<do stuff that updates state>>
}
In Haskell we're obviously not going to be checking a variable against true as the loop condition, because it can't change its value[1] and we'd either evaluate the loop body forever or never. So instead, we'll want to be evaluating a function that returns a boolean value on some argument:
while (check something == True) {
<<do stuff that updates state>>
}
Well, now we don't have a state variable, so that "do stuff that updates state" is looking pretty pointless. And we don't have a something to pass to check. Lets think about this a bit more. We want the something to be checked to depend on what the "do stuff" bit is doing. We don't have side effects, so that means something has to be (or be derived from) returned from the "do stuff". "do stuff" also needs to take something that varies as an argument, or it'll just keep returning the same thing forever, which is also pointless. We also need to return a value out all this, otherwise we're just burning CPU cycles (again, with no side effects there's no point running a function if we don't use its output in some way, and there's even less point running a function repeatedly if we never use its output).
So how about something like this:
while check func state =
let next_state = func state in
if check next_state
then while check func next_state
else next_state
Lets try it in GHCi:
*Main> while (<20) (+1) 0
20
This is the result of applying (+1) repeatedly while the result is less than 20, starting from 0.
*Main> while ((<20) . length) (++ "zob") ""
"zobzobzobzobzobzobzob"
This is the result of concatenating "zob" repeatedly while the result's length is less than 20, starting from the empty string.
So you can see I've defined a function that is (sort of a bit) analogous to a while loop from imperative languages. We didn't even need dedicated loop syntax for it! (which is the real reason Haskell has no such syntax; if you need this kind of thing you can express it as a function). It's not the only way to do so, and experienced Haskell programmers would probably use other standard library functions to do this kind of job, rather than writing while.
But I think it's useful to see how you can express this kind of thing in Haskell. It does show that you can't translate things like imperative loops directly into Haskell; I didn't end up translating your loop in terms of my while because it ends up pretty pointless; you never use the result of function1 or function2, they're called with no arguments so they'd always return the same thing in every iteration, and function3 likewise always returns the same thing, and can only return true or false to either cause while to keep looping or stop, with no information resulting.
Presumably in the C++ program they're all using side effects to actually get some work done. If they operate on in-memory things then you need to translate a bigger chunk of your program at once to Haskell for the translation of this loop to make any sense. If those functions are doing IO then you'll need to do this in the IO monad in Haskell, for which my while function doesn't work, but you can do something similar.
[1] As an aside, it's worth trying to understand that "you can't change variables" in Haskell isn't just an arbitrary restriction, nor is it just an acceptable trade off for the benefits of purity, it is a concept that doesn't make sense the way Haskell wants you to think about Haskell code. You're writing down expressions that result from evaluating functions on certain arguments: in f x = x + 1 you're saying that f x is x + 1. If you really think of it that way rather than thinking "f takes x, then adds one to it, then returns the result" then the concept of "having side effects" doesn't even apply; how could something existing and being equal to something else somehow change a variable, or have some other side effect?
You should write a solution to your problem in a more functional approach.
However, some code in haskell works a lot like imperative looping, take for example state monads, terminal recursivity, until, foldr, etc.
A simple example is the factorial. In C, you would write a loop where in haskell you can for example write fact n = foldr (*) 1 [2..n].
If you've two functions f :: a -> b and g :: b -> c where a, b, and c are types like String or [Int] then you can compose them simply by writing f . b.
If you wish them to loop over a list or vector you could write map (f . g) or V.map (f . g), assuming you've done Import qualified Data.Vector as V.
Example : I wish to print a list of markdown headings like ## <number>. <heading> ## but I need roman numerals numbered from 1 and my list headings has type type [(String,Double)] where the Double is irrelevant.
Import Data.List
Import Text.Numeral.Roman
let fun = zipWith (\a b -> a ++ ". " ++ b ++ "##\n") (map toRoman [1..]) . map fst
fun [("Foo",3.5),("Bar",7.1)]
What the hell does this do?
toRoman turns a number into a string containing the roman numeral. map toRoman does this to every element of a loop. map toRoman [1..] does it to every element of the lazy infinite list [1,2,3,4,..], yielding a lazy infinite list of roman numeral strings
fst :: (a,b) -> a simply extracts the first element of a tuple. map fst throws away our silly Meow information along the entire list.
\a b -> "##" ++ show a ++ ". " ++ b ++ "##" is a lambda expression that takes two strings and concatenates them together within the desired formatting strings.
zipWith :: (a -> b -> c) -> [a] -> [b] -> [c] takes a two argument function like our lambda expression and feeds it pairs of elements from it's own second and third arguments.
You'll observe that zip, zipWith, etc. only read as much of the lazy infinite list of Roman numerals as needed for the list of headings, meaning I've number my headings without maintaining any counter variable.
Finally, I have declared fun without naming it's argument because the compiler can figure it out from the fact that map fst requires one argument. You'll notice that put a . before my second map too. I could've written (map fst h) or $ map fst h instead if I'd written fun h = ..., but leaving the argument off fun meant I needed to compose it with zipWith after applying zipWith to two arguments of the three arguments zipWith wants.
I'd hope the compiler combines the zipWith and maps into one single loop via inlining.
(I'm sure this must have been answered on this site already, but search gets inundated with the concept of calling free() on a variable in C.)
I came across the term "eta reduction," which was defined something like f x = M x ==> M if x is "not free in M". I mean, I think I understand the gist of what it's trying to say, it seems like what you do when you convert a function to point-free style, but I don't know what the qualifier about x not being free means.
Here's an example:
\f -> f x
In this lambda, x is a free variable. Basically a free variable is a variable used in a lambda that is not one of the lambda's arguments (or a let variable). It comes from outside the context of the lambda.
Eta reduction means we can change:
(\x -> g x) to (g)
But only if x is not free (i.e. it is not used or is an argument) in g. Otherwise we'd be creating an expression which refers to a unknown variable:
(\x -> (x+) x) to (x+) ???
Well, here's the relevant Wikipedia article, for what that's worth.
The short version is that such definitions elide the body of a lambda expression using a placeholder like "M", and so have to specify additionally that the variable being bound by that lambda isn't used in whatever the placeholder represents.
So, a "free variable" here roughly means a variable defined in some ambiguous or unknown outer scope--e.g., in an expression like \y -> x + y, x is a free variable but y is not.
Eta reduction is about removing a superfluous layer of binding and immediately applying a variable, which is (as you would probably imagine) only valid if the variable in question is only used in that one place.