I'am quite new to F#, and was solving some basic exercises when i stumbled upon this function
Give the (most general) types of g1 and g2 and describe what each of these two functions
computes. Your description for each function should focus on what it computes, rather
than on individual computation steps
let rec g1 p = function
| x::xs when p x -> x :: g1 p xs
| _ -> [];;
i don't the understand " when p x " part, or how to call the function. can someone please explain what this function takes in as an argument? as just calling the function like that " g1 [1;2;3] " gives me an error.
Tried calling the function, and tried reading some text books to figure it out
The function keyword is a bit tricky, but it's just syntactical sugar for a match expression. The following two functions are equivalent:
let fooMatch x =
match x with
| 1 -> "one"
| _ -> "not one"
let fooFunction =
function
| 1 -> "one"
| _ -> "not one"
If you use function instead of match, then the (last) argument to the function becomes implicit, rather than explicit, but it's still there. Both versions of foo take one argument.
The when p x -> part is called a guard. In your case, p stands for "predicate" (a function that returns true/false). To call your function, you need pass both a predicate and a list. E.g. g1 (fun x -> x % 2 = 0) [1;2;3].
so this is clearly homework, so I don't want to answer it in full just give guidance.
let rec g1 p = function
| x::xs when p x -> x :: g1 p xs
| _ -> [];;
is tricky because "function" is sort of shorthand sort of hides a parameter, the above is the same as (well almost, but for your purposes it is).
let rec g1 p ys =
match ys with
| x::xs when p x -> x :: g1 p xs
| _ -> [];;
so you can immediately see g1 is a function that takes 2 parameters and returns something
g1 : ? -> ? -> ?
you just need to try and work out if there is anything more you can infer anythng about the ?s.
So F# will look at your implementation and try and infer what it can.
as said by brian, "when p x" is a guard, so p is a function that takes an x and returns a bool
p : ? -> bool
and p is the first param of g1 so
g1 : (? -> bool) -> ? -> ?
etc.
(just be aware you/the F# compiler may not be able to identify a concrete type, like 'bool', sometimes it will infer that it doesnt know the specific type i.e. the function is has a 'parametric' type (generic), so you can get types like
identityFunction : 'a -> 'a
i.e. whatever the type of the thing being passed in, something of the same type will be returned
2nd note is, this function is recursive, so you have to make sure that the types in the recursive call inside the function line up with the types of the function you infer by inspecting the code)
Related
I'm learning F# and I cannot figure out what the difference between let, fun and function is, and my text book doesn't really explain that either. As an example:
let s sym = function
| V x -> Map.containsKey x sym
| A(f, es) -> Map.containsKey f sym && List.forall (s sym) es;;
Couldn't I have written this without the function keyword? Or could I have written that with fun instead of function? And why do I have to write let when I've seen some examples where you write
fun s x =
...
What's the difference really?
I guess you should really ask MSDN, but in a nutshell:
let binds a value with a symbol. The value can be a plain type like an int or a string, but it can also be a function. In FP functions are values and can be treated in the same way as those types.
fun is a keyword that introduces an anonymous function - think lambda expression if you're familiar with C#.
Those are the two important ones, in the sense that all the others usages you've seen can be thought as syntax sugar for those two. So to define a function, you can say something like this:
let myFunction =
fun firstArg secondArg ->
someOperation firstArg secondArg
And that's very clear way of saying it. You declare that you have a function and then bind it to the myFunction symbol.
But you can save yourself some typing by just conflating anonymous function declaration and binding it to a symbol with let:
let myFunction firstArg secondArg =
someOperation firstArg secondArg
What function does is a bit trickier - you combine an anonymous single-argument function declaration with a match expression, by matching on an implicit argument. So these two are equivalent:
let myFunction firstArg secondArg =
match secondArg with
| "foo" -> firstArg
| x -> x
let myFunction firstArg = function
| "foo" -> firstArg
| x -> x
If you're just starting on F#, I'd steer clear of that one. It has its uses (mainly for providing succinct higher order functions for maps/filters etc.), but results in code less readable at a glance.
These things are sort of shortcuts to each other.
The most fundamental thing is let. This keyword gives names to stuff:
let name = "stuff"
Speaking more technically, the let keyword defines an identifier and binds it to a value:
let identifier = "value"
After this, you can use words name and identifier in your program, and the compiler will know what they mean. Without the let, there wouldn't be a way to name stuff, and you'd have to always write all your stuff inline, instead of referring to chunks of it by name.
Now, values come in different flavors. There are strings "some string", there are integer numbers 42, floating point numbers 5.3, Boolean values true, and so on. One special kind of value is function. Functions are also values, in most respects similar to strings and numbers. But how do you write a function? To write a string, you use double quotes, but what about function?
Well, to write a function, you use the special word fun:
let squareFn = fun x -> x*x
Here, I used the let keyword to define an identifier squareFn, and bind that identifier to a value of the function kind. Now I can use the word squareFn in my program, and the compiler will know that whenever I use it I mean a function fun x -> x*x.
This syntax is technically sufficient, but not always convenient to write. So in order to make it shorter, the let binding takes an extra responsibility upon itself and provides a shorter way to write the above:
let squareFn x = x*x
That should do it for let vs fun.
Now, the function keyword is just a short form for fun + match. Writing function is equivalent to writing fun x -> match x with, period.
For example, the following three definitions are equivalent:
let f = fun x ->
match x with
| 0 -> "Zero"
| _ -> "Not zero"
let f x = // Using the extra convenient form of "let", as discussed above
match x with
| 0 -> "Zero"
| _ -> "Not zero"
let f = function // Using "function" instead of "fun" + "match"
| 0 -> "Zero"
| _ -> "Not zero"
I'm looking through some notes that my professor gave regarding the language SML and one of the functions looks like this:
fun max gt =
let fun lp curr [] = curr
| lp curr (a::l) = if gt(a,curr)
then lp a l
else lp curr l
in
lp
end
Could someone help explain what this is doing? The thing that I am most confused about is the line:
let fun lp curr [] = curr
What exactly does this mean? As far as I can tell there is a function called lp but what does the curr [] mean? Are these arguments? If so, aren't you only allowed one parameter in sml?
It means that lp is a function that takes 2 parameters, the first being curr and the second being, well, a list, which logically, may be either empty ([]) or contain at least one element ((a::l) is a pattern for a list where a is at the head, and the rest of the list is l).
If one were to translate that bit of FP code into a certain well-known imperative language, it would look like:
function lp(curr, lst) {
if (lst.length == 0) {
return curr;
} else {
var a = lst[0]; // first element
var l = lst.slice(1, lst.length); // the rest
if (gt(a, curr)) {
return lp(a, l);
} else {
return lp(curr, l)
}
}
}
Quite a mouthful, but it's a faithful translation.
Functional languages are based on the Lambda Calculus, where functions take exactly one value and return one result. While SML and other FP languages are based on this theory, it's rather inconvenient in practice, so many of these languages allow you to express passing multiple parameters to a function via what is known as Currying.
So yes, in ML functions actually take only one value, but currying lets you emulate multiple arguments.
Let's create a function called add, which adds 2 numbers:
fun add a b = a + b
should do it, but we defined 2 parameters. What's the type of add? If you take a look in the REPL, it is val add = fn : int -> int -> int. Which reads, "add is a function that takes an int and returns another function (which takes an int and returns an int)"
So we could also have defined add this way:
fun add a =
fn b => a + b
And you will see that they are alike. In fact it is safe to say that in a way,
the former is syntactic sugar for the later.
So all functions you define in ML, even those with several arguments, are actually functions with one argument, that return functions that accept the second argument and so on. It's a little hard to get used to at first but it
becomes second nature very soon.
fun add a b = a + b (* add is of type int -> int -> int *)
add 1 2 (* returns 3 as you expect *)
(* calling add with only one parameter *)
val add1 = add 1
What's add1? It is a function that will add 1 to the single argument you pass it!
add1 2 (* returns 3 *)
This is an example of partial application, where you are calling a function piecemeal,
one argument at a time, getting back each time, another function that accepts the rest
of the arguments.
Also, there's another way to give the appearance of multiple arguments: tuples:
(1, 2); (* evaluates to a tuple of (int,int) *)
fun add (a,b) = a + b;
add (1, 2) (* passing a SINGLE argument to a function that
expects only a single argument, a tuple of 2 numbers *)
In your question, lp could have also been implemented as lp (curr, someList):
fun max gt curr lst =
let fun lp (curr, []) = curr
| lp (curr, (a::l)) = if gt(a,curr) then lp (a, l)
else lp (curr, l)
in
lp (curr, lst)
end
Note that in this case, we have to declare max as max gt curr lst!
In the code you posted, lp was clearly implemented with currying. And the type of
max itself was fn: ('a * 'a -> bool) -> 'a -> 'a list -> 'a. Taking that apart:
('a * 'a -> bool) -> (* passed to 'max' as 'gt' *)
'a -> (* passed to 'lp' as 'curr' *)
'a list -> (* passed to 'lp' as 'someList' *)
'a (* what 'lp' returns (same as what 'max' itself returns) *)
Note the type of gt, the first argument to max: fn : (('a * 'a) -> bool) - it is a function of one argument ('a * 'a), a tuple of two 'a's and it returns an 'a. So no currying here.
Which to use is a matter of both taste, convention and practical considerations.
Hope this helps.
Just to clarify a bit on currying, from Faiz's excellent answer.
As previously stated SML only allows functions to take 1 argument. The reason for this is because a function fun foo x = x is actually a derived form of (syntactic sugar) val rec foo = fn x => x. Well actually this is not entirely true, but lets keep it simple for a second
Now take for example this power function. Here we declare the function to "take two arguments"
fun pow n 0 = 1
| pow n k = n * pow n (k-1)
As stated above, fun ... was a derived form and thus the equivalent form of the power function is
val rec pow = fn n => fn k => ...
As you might see here, we have a problem expressing the two different pattern matches of the original function declaration, and thus we can't keep it "simple" anymore and the real equivalent form of a fun declaration is
val rec pow = fn n => fn k =>
case (n, k) of
(n, 0) => 1
| (n, k) => n * pow n (k-1)
For the sake of completeness, cases is actually also a derived form, with anonymous functions as the equivalent form
val rec pow = fn n => fn k =>
(fn (n,0) => 1
| (n,k) => n * pow n (k-1)) (n,k)
Note that (n,k) is applied directly to the inner most anonymous function.
"Write a function lv: cfg -> (blabel -> ide set), which computes the live variables analysis on the given control flow graph."
Having cfg and blabel defined and ide set as a list of string, how can I create a function with that signature?
You're presumably familiar with the let syntax to define a function:
let f x = x + 1 in …
You can use this syntax anywhere, including in a function body. Now if you happen to use the inner function's name as the return value of the outer function, the outer function will be returning a function.
let outer_function x =
let inner_function y = x + y in
inner_function
The let syntax is in fact syntactic sugar for fun or function. In particular, if you define inner_function just to use the name once, you might as well use the fun notation and not bother giving the inner function a name.
let outer_function x =
fun y -> x + y
Furthermore, if all the outer function does when you pass it an argument is to build and return an inner function, then consider its behavior when you pass that function two arguments. First the outer function builds an inner function, using its first (and only) argument; then that inner function is applied to the second argument, so its body is executed. This is equivalent to having just one function that takes two arguments. This
observation is known as currying.
let outer_function x y = x + y
Note that the type of this function is int -> int -> int; this is the same type as int -> (int -> int) (the arrow type operator is right-associative).
Currying doesn't apply when the outer function does some work before building the inner function. In that case, the work is performed after receiving the first argument.
let outer_function x =
print_endline "outer";
fun y -> print_endline "inner"; x + y
So the structure of your code is likely to look like this:
let lv (c : cfg) : blabel -> ide set =
let c' = do_some_precomputation c in
fun (bl : blabel) -> (… : ide set)
Do anyone can give example of function composition?
This is the definition of function composition operator?
(.) :: (b -> c) -> (a -> b) -> a -> c
f . g = \x -> f (g x)
This shows that it takes two functions and return a function but i remember someone has expressed the logic in english like
boy is human -> ali is boy -> ali is human
What this logic related to function composition?
What is the meaning of strong binding of function application and composition and which one is more strong binding than the other?
Please help.
Thanks.
(Edit 1: I missed a couple components of your question the first time around; see the bottom of my answer.)
The way to think about this sort of statement is to look at the types. The form of argument that you have is called a syllogism; however, I think you are mis-remembering something. There are many different kinds of syllogisms, and yours, as far as I can tell, does not correspond to function composition. Let's consider a kind of syllogism that does:
If it is sunny out, I will get hot.
If I get hot, I will go swimming.
Therefore, if it is sunny about, I will go swimming.
This is called a hypothetical syllogism. In logic terms, we would write it as follows: let S stand for the proposition "it is sunny out", let H stand for the proposition "I will get hot", and let W stand for the proposition "I will go swimming". Writing α → β for "α implies β", and writing ∴ for "therefore", we can translate the above to:
S → H
H → W
∴ S → W
Of course, this works if we replace S, H, and W with any arbitrary α, β, and γ. Now, this should look familiar. If we change the implication arrow → to the function arrow ->, this becomes
a -> b
b -> c
∴ a -> c
And lo and behold, we have the three components of the type of the composition operator! To think about this as a logical syllogism, you might consider the following:
If I have a value of type a, I can produce a value of type b.
If I have a value of type b, I can produce a value of type c.
Therefore, if I have a value of type a, I can produce a value of type c.
This should make sense: in f . g, the existence of a function g :: a -> b tells you that premise 1 is true, and f :: b -> c tells you that premise 2 is true. Thus, you can conclude the final statement, for which the function f . g :: a -> c is a witness.
I'm not entirely sure what your syllogism translates to. It's almost an instance of modus ponens, but not quite. Modus ponens arguments take the following form:
If it is raining, then I will get wet.
It is raining.
Therefore, I will get wet.
Writing R for "it is raining", and W for "I will get wet", this gives us the logical form
R → W
R
∴ W
Replacing the implication arrow with the function arrow gives us the following:
a -> b
a
∴ b
And this is simply function application, as we can see from the type of ($) :: (a -> b) -> a -> b. If you want to think of this as a logical argument, it might be of the form
If I have a value of type a, I can produce a value of type b.
I have a value of type a.
Therefore, I can produce a value of type b.
Here, consider the expression f x. The function f :: a -> b is a witness of the truth of proposition 1; the value x :: a is a witness of the truth of proposition 2; and therefore the result can be of type b, proving the conclusion. It's exactly what we found from the proof.
Now, your original syllogism takes the following form:
All boys are human.
Ali is a boy.
Therefore, Ali is human.
Let's translate this to symbols. Bx will denote that x is a boy; Hx will denote that x is human; a will denote Ali; and ∀x. φ says that φ is true for all values of x. Then we have
∀x. Bx → Hx
Ba
∴ Ha
This is almost modus ponens, but it requires instantiating the forall. While logically valid, I'm not sure how to interpret it at the type-system level; if anybody wants to help out, I'm all ears. One guess would be a rank-2 type like (forall x. B x -> H x) -> B a -> H a, but I'm almost sure that that's wrong. Another guess would be a simpler type like (B x -> H x) -> B Int -> H Int, where Int stands for Ali, but again, I'm almost sure it's wrong. Again: if you know, please let me know too!
And one last note. Looking at things this way—following the connection between proofs and programs—eventually leads to the deep magic of the Curry-Howard isomorphism, but that's a more advanced topic. (It's really cool, though!)
Edit 1: You also asked for an example of function composition. Here's one example. Suppose I have a list of people's middle names. I need to construct a list of all the middle initials, but to do that, I first have to exclude every non-existent middle name. It is easy to exclude everyone whose middle name is null; we just include everyone whose middle name is not null with filter (\mn -> not $ null mn) middleNames. Similarly, we can easily get at someone's middle initial with head, and so we simply need map head filteredMiddleNames in order to get the list. In other words, we have the following code:
allMiddleInitials :: [Char]
allMiddleInitials = map head $ filter (\mn -> not $ null mn) middleNames
But this is irritatingly specific; we really want a middle-initial–generating function. So let's change this into one:
getMiddleInitials :: [String] -> [Char]
getMiddleInitials middleNames = map head $ filter (\mn -> not $ null mn) middleNames
Now, let's look at something interesting. The function map has type (a -> b) -> [a] -> [b], and since head has type [a] -> a, map head has type [[a]] -> [a]. Similarly, filter has type (a -> Bool) -> [a] -> [a], and so filter (\mn -> not $ null mn) has type [a] -> [a]. Thus, we can get rid of the parameter, and instead write
-- The type is also more general
getFirstElements :: [[a]] -> [a]
getFirstElements = map head . filter (not . null)
And you see that we have a bonus instance of composition: not has type Bool -> Bool, and null has type [a] -> Bool, so not . null has type [a] -> Bool: it first checks if the given list is empty, and then returns whether it isn't. This transformation, by the way, changed the function into point-free style; that is, the resulting function has no explicit variables.
You also asked about "strong binding". What I think you're referring to is the precedence of the . and $ operators (and possibly function application as well). In Haskell, just as in arithmetic, certain operators have higher precedence than others, and thus bind more tightly. For instance, in the expression 1 + 2 * 3, this is parsed as 1 + (2 * 3). This is because in Haskell, the following declarations are in force:
infixl 6 +
infixl 7 *
The higher the number (the precedence level), the sooner that that operator is called, and thus the more tightly the operator binds. Function application effectively has infinite precedence, so it binds the most tightly: the expression f x % g y will parse as (f x) % (g y) for any operator %. The . (composition) and $ (application) operators have the following fixity declarations:
infixr 9 .
infixr 0 $
Precedence levels range from zero to nine, so what this says is that the . operator binds more tightly than any other (except function application), and the $ binds less tightly. Thus, the expression f . g $ h will parse as (f . g) $ h; and in fact, for most operators, f . g % h will be (f . g) % h and f % g $ h will be f % (g $ h). (The only exceptions are the rare few other zero or nine precedence operators.)
This question here is related to
Haskell Input Return Tuple
I wonder how we can passes the input from monad IO to another function in order to do some computation.
Actually what i want is something like
-- First Example
test = savefile investinput
-- Second Example
maxinvest :: a
maxinvest = liftM maximuminvest maxinvestinput
maxinvestinput :: IO()
maxinvestinput = do
str <- readFile "C:\\Invest.txt"
let cont = words str
let mytuple = converttuple cont
let myint = getint mytuple
putStrLn ""
-- Convert to Tuple
converttuple :: [String] -> [(String, Integer)]
converttuple [] = []
converttuple (x:y:z) = (x, read y):converttuple z
-- Get Integer
getint :: [(String, Integer)] -> [Integer]
getint [] = []
getint (x:xs) = snd (x) : getint xs
-- Search Maximum Invest
maximuminvest :: (Ord a) => [a] -> a
maximuminvest [] = error "Empty Invest Amount List"
maximuminvest [x] = x
maximuminvest (x:xs)
| x > maxTail = x
| otherwise = maxTail
where maxTail = maximuminvest xs
In the second example, the maxinvestinput is read from file and convert the data to the type maximuminvest expected.
Please help.
Thanks.
First, I think you're having some basic issues with understanding Haskell, so let's go through building this step by step. Hopefully you'll find this helpful. Some of it will just arrive at the code you have, and some of it will not, but it is a slowed-down version of what I'd be thinking about as I wrote this code. After that, I'll try to answer your one particular question.
I'm not quite sure what you want your program to do. I understand that you want a program which reads as input a file containing a list of people and their investments. However, I'm not sure what you want to do with it. You seem to (a) want a sensible data structure ([(String,Integer)]), but then (b) only use the integers, so I'll suppose that you want to do something with the strings too. Let's go through this. First, you want a function that can, given a list of integers, return the maximum. You call this maximuminvest, but this function is more general that just investments, so why not call it maximum? As it turns out, this function already exists. How could you know this? I recommend Hoogle—it's a Haskell search engine which lets you search both function names and types. You want a function from lists of integers to a single integer, so let's search for that. As it turns out, the first result is maximum, which is the more general version of what you want. But for learning purposes, let's suppose you want to write it yourself; in that case, your implementation is just fine.
Alright, now we can compute the maximum. But first, we need to construct our list. We're going to need a function of type [String] -> [(String,Integer)] to convert our formattingless list into a sensible one. Well, to get an integer from a string, we'll need to use read. Long story short, your current implementation of this is also fine, though I would (a) add an error case for the one-item list (or, if I were feeling nice, just have it return an empty list to ignore the final item of odd-length lists), and (b) use a name with a capital letter, so I could tell the words apart (and probably a different name):
tupledInvestors :: [String] -> [(String, Integer)]
tupledInvestors [] = []
tupledInvestors [_] = error "tupledInvestors: Odd-length list"
tupledInvestors (name:amt:rest) = (name, read amt) : tupledInvestors rest
Now that we have these, we can provide ourselves with a convenience function, maxInvestment :: [String] -> Integer. The only thing missing is the ability to go from the tupled list to a list of integers. There are several ways to solve this. One is the one you have, though that would be unusual in Haskell. A second would be to use map :: (a -> b) -> [a] -> [b]. This is a function which applies a function to every element of a list. Thus, your getint is equivalent to the simpler map snd. The nicest way would probably be to use Data.List.maximumBy :: :: (a -> a -> Ordering) -> [a] -> a. This is like maximum, but it allows you to use a comparison function of your own. And using Data.Ord.comparing :: Ord a => (b -> a) -> b -> b -> Ordering, things become nice. This function allows you to compare two arbitrary objects by converting them to something which can be compared. Thus, I would write
maxInvestment :: [String] -> Integer
maxInvestment = maximumBy (comparing snd) . tupledInvestors
Though you could also write maxInvestment = maximum . map snd . tupledInvestors.
Alright, now on to the IO. Your main function, then, wants to read from a specific file, compute the maximum investment, and print that out. One way to represent that is as a series of three distinct steps:
main :: IO ()
main = do dataStr <- readFile "C:\\Invest.txt"
let maxInv = maxInvestment $ words dataStr
print maxInv
(The $ operator, if you haven't seen it, is just function application, but with more convenient precedence; it has type (a -> b) -> a -> b, which should make sense.) But that let maxInv seems pretty pointless, so we can get rid of that:
main :: IO ()
main = do dataStr <- readFile "C:\\Invest.txt"
print . maxInvestment $ words dataStr
The ., if you haven't seen it yet, is function composition; f . g is the same as \x -> f (g x). (It has type (b -> c) -> (a -> b) -> a -> c, which should, with some thought, make sense.) Thus, f . g $ h x is the same as f (g (h x)), only easier to read.
Now, we were able to get rid of the let. What about the <-? For that, we can use the =<< :: Monad m => (a -> m b) -> m a -> m b operator. Note that it's almost like $, but with an m tainting almost everything. This allows us to take a monadic value (here, the readFile "C:\\Invest.txt" :: IO String), pass it to a function which turns a plain value into a monadic value, and get that monadic value. Thus, we have
main :: IO ()
main = print . maxInvestment . words =<< readFile "C:\\Invest.txt"
That should be clear, I hope, especially if you think of =<< as a monadic $.
I'm not sure what's happening with testfile; if you edit your question to reflect that, I'll try to update my answer.
One more thing. You said
I wonder how we can passes the input from monad IO to another function in order to do some computation.
As with everything in Haskell, this is a question of types. So let's puzzle through the types here. You have some function f :: a -> b and some monadic value m :: IO a. You want to use f to get a value of type b. This is impossible, as I explained in my answer to your other question; however, you can get something of type IO b. Thus, you need a function which takes your f and gives you a monadic version. In other words, something with type Monad m => (a -> b) -> (m a -> m b). If we plug that into Hoogle, the first result is Control.Monad.liftM, which has precisely that type signature. Thus, you can treat liftM as a slightly different "monadic $" than =<<: f `liftM` m applies f to the pure result of m (in accordance with whichever monad you're using) and returns the monadic result. The difference is that liftM takes a pure function on the left, and =<< takes a partially-monadic one.
Another way to write the same thing is with do-notation:
do x <- m
return $ f x
This says "get the x out of m, apply f to it, and lift the result back into the monad." This is the same as the statement return . f =<< m, which is precisely liftM again. First f performs a pure computation; its result is passed into return (via .), which lifts the pure value into the monad; and then this partially-monadic function is applied, via =<,, to m.
It's late, so I'm not sure how much sense that made. Let me try to sum it up. In short, there is no general way to leave a monad. When you want to perform computation on monadic values, you lift pure values (including functions) into the monad, and not the other way around; that could violate purity, which would be Very Bad™.
I hope that actually answered your question. Let me know if it didn't, so I can try to make it more helpful!
I'm not sure I understand your question, but I'll answer as best I can. I've simplified things a bit to get at the "meat" of the question, if I understand it correctly.
maxInvestInput :: IO [Integer]
maxInvestInput = liftM convertToIntegers (readFile "foo")
maximumInvest :: Ord a => [a] -> a
maximumInvest = blah blah blah
main = do
values <- maxInvestInput
print $ maximumInvest values
OR
main = liftM maximumInvest maxInvestInput >>= print