Haskell function definition and arguments list - function

I have the following function definitions (which are equivalent):
reverseWords :: String -> String
reverseWords wds = unwords . map reverse . words $ wds
and
reverseWords :: String -> String
reverseWords = unwords . map reverse . words
I understand that they are equivalent as they yield the same result but I'm a bit confused about the second form.
If I call a function like:
reverseWords abc
How does haskell decide where to place this 'abc' parameter?

Haskell does not need to decide where to place abc. When you call reverseWords in your second example it will simply return unwords . map reverse . words.
And then abc gets applied to that with which you will end up with the same result as in your first example.
This is also called Point-free style as you can read more about it here: https://wiki.haskell.org/Point-free

Related

How to partially apply a function with desired order in Elm?

Suppose I have a function that takes 3 parameters as input. How to partially apply this function in Elm so it takes first and last parameters and waits for the second parameter to return the final result?
This can be done in Ramda with R.__ which is named placeholer.
You can just wrap it in a lambda function that has the shape you want, which is what would be produced by any other means anyway:
\y -> f "x" y "z"
In a curried language I find the need to do this so rare that adding syntax sugar specifically for this use case seems unnecessary.
As glennsl says, you can wrap your function in another function with the argument order that you want. his answer assumes that you know statically what the first and third arguments are, if you don't, but just want to partially apply the first and third argument, then apply the second you can take a function like,
joinThree : String -> String -> String -> String
joinThree first second third =
first ++ second ++ third
and wrap it in a new function that calls the first function, but with a different argument order,
joinThreeWrapper : String -> String -> String -> String
joinThreeWrapper first third second =
joinThree first second third
This allows you to call this function like,
welcomeToNeverland : String -> String
welcomeToNeverland name =
let
myGreeting = joinThreeWrapper "Welcome " " to Neverland"
in
myGreeting name
Then you can use it like,
text (welcomeToNeverland "Wendy")
-- Welcome Wendy to Neverland
Writing joinThreeWrapper like this makes it a bit easier to map your function over a list like,
greetMany : List String -> List String
greetMany names =
List.map (joinThreeWrapper "Welcome " ", this is our town. ") names
so that you can do,
text (List.map (++) (greetMany ["Jesse", "Carl"]))
-- Welcome Jesse, this is our town. Welcome Carl, this is our town.
You can use flip from the core Basics module.
For instance:
> append3 x y z = x ++ y ++ z
<function> : appendable -> appendable -> appendable -> appendable
> hello = flip (append3 "Hello, ") "!"
<function> : String -> String
> hello "world"
"Hello, world!" : String

How do I pass objects to functions in ooRexx?

I'm a long-time mainframe Rexx programmer who is trying out objects in ooRexx. The results are surprising. For example, here is a program:
#!/usr/bin/rexx
a = .number~new(3.14)
say "a =" a
say "a~val =" a~val
call say_number a
exit 0
say_number:
procedure
parse arg num
say "In say_number"
say "num =" num
say "num~val =" num~val
return
::class number public
::attribute val get public
::method init ; expose val ; use arg val
::method new ; expose val ; use arg val
::method string ; return "'"self~val"'"
The result is:
> number
a = '3.14'
a~val = 3.14
In say_number
num = '3.14'
18 *-* say "num~val =" num~val
8 *-* call say_number a
REX0097E: Error 97 running /home/tony/bin/.scripts/number line 18: Object method not found
REX0476E: Error 97.1: Object "'3.14'" does not understand message "VAL"
It appears that the object is being resolved to its string value before it's passed to the say_number function. Weird! Am I missing something obvious?
Well, that didn't take long. I changed parse to use in the function, and everything worked as expected. Per the Reference manual:
USE ARG retrieves the argument objects provided in a program, routine,
function, or method and assigns them to variables or message term
assignments.
PARSE assigns data from various sources to one or more variables
according to the rules of parsing. ... If you specify UPPER, the
strings to be parsed are translated to uppercase before parsing. If
you specify LOWER, the strings are translated to lowercase. Otherwise
no translation takes place.
Presumably PARSE converts the arguments to a string so that it can change case as requested (or defaulted).

Haskell - have a function call a function

we are currently sitting on a task from university, which we don't fully understand (please no solution but only ideas or suggestions).
What is given is a type:
type MyType = String -> String
Now we are trying to be able to have a function, which takes 2 Strings and a function (the type) and then gives a function (type)
myCode :: String -> String -> MyType -> MyType
and we already implemented a function, which can be used as MyType one:
emptyString :: MyType
emptyString :: (\a -> "")
The task is to be able to store several 2x Strings. This is our current idea:
myCode :: String -> String -> MyType ->MyType
myCode a b c = (\x -> b)
in this case we have an input String, which is "Hello" and another one which is "World" and then as c we put in the "emptyString".
This works for one String, because when we type the following in the console:
a = (myCode "Hello" "World" emptyString) ""
we get "World" on input "a". Now the hard part: We should be able to store several of these (searching them is another task, not needed right now). We thought we might be able to use "a" now when declaring another variable:
b = (myCode "1" "2" a) "Hello" "World" emptyString "")
This would call in "b" the function saved as "a" and within this the "emptyString".
As you may have guessed - it doesn't work! And we are really at a loss on how to carry on from now.
When you reached this part, it means you took the time to understand our complicated explanation of our task - thanks a lot.
Thanks for suggestions and help in advance!
From the question linked by amalloy in the comments, it looks like you are trying to build a phonebook based on a continuation passing style like paradigm.
Basically, what is supposed to happen for your type
myCode :: String -> String -> MyType -> MyType
is that you will generate a piece of data dat = myCode a b pb, which is of type MyType. So, you can query dat with an s :: String and it will output another String. In the operation of dat s, if you expand it to the definition,
dat s = myCode a b pb s
you have access to three strings, a, b, and whatever pb s returns. You will build up functionality recursively, either by doing something with a b and s, or pushing it down the road to pb, letting the continuation handle it.
Hope this helps without giving too much away.

Confused about this nested function

I am reading the Python Cookbook 3rd Edition and came across the topic discussed in 2.6 "Searching and Replacing Case-Insensitive Text," where the authors discuss a nested function that is like below:
def matchcase(word):
def replace(m):
text = m.group()
if text.isupper():
return word.upper()
elif text.islower():
return word.lower()
elif text[0].isupper():
return word.capitalize()
else:
return word
return replace
If I have some text like below:
text = 'UPPER PYTHON, lower python, Mixed Python'
and I print the value of 'text' before and after, the substitution happens correctly:
x = matchcase('snake')
print("Original Text:",text)
print("After regsub:", re.sub('python', matchcase('snake'), text, flags=re.IGNORECASE))
The last "print" command shows that the substitution correctly happens but I am not sure how this nested function "gets" the:
PYTHON, python, Python
as the word that needs to be substituted with:
SNAKE, snake, Snake
How does the inner function replace get its value 'm'?
When matchcase('snake') is called, word takes the value 'snake'.
Not clear on what the value of 'm' is.
Can any one help me understand this clearly, in this case?
Thanks.
When you pass a function as the second argument to re.sub, according to the documentation:
it is called for every non-overlapping occurrence of pattern. The function takes a single match object argument, and returns the replacement string.
The matchcase() function itself returns the replace() function, so when you do this:
re.sub('python', matchcase('snake'), text, flags=re.IGNORECASE)
what happens is that matchcase('snake') returns replace, and then every non-overlapping occurrence of the pattern 'python' as a match object is passed to the replace function as the m argument. If this is confusing to you, don't worry; it is just generally confusing.
Here is an interactive session with a much simpler nested function that should make things clearer:
In [1]: def foo(outer_arg):
...: def bar(inner_arg):
...: print(outer_arg + inner_arg)
...: return bar
...:
In [2]: f = foo('hello')
In [3]: f('world')
helloworld
So f = foo('hello') is assigning a function that looks like the one below to a variable f:
def bar(inner_arg):
print('hello' + inner_arg)
f can then be called like this f('world'), which is like calling bar('world'). I hope that makes things clearer.

Partial application of operators

If I want to add a space at the end of a character to return a list, how would I accomplish this with partial application if I am passing no arguments?
Also would the type be?
space :: Char -> [Char]
I'm having trouble adding a space at the end due to a 'parse error' by using the ++ and the : operators.
What I have so far is:
space :: Char -> [Char]
space = ++ ' '
Any help would be much appreciated! Thanks
Doing what you want is so common in Haskell it's got its own syntax, but being Haskell, it's extraordinarily lightweight. For example, this works:
space :: Char -> [Char]
space = (:" ")
so you weren't far off a correct solution. ([Char] is the same as String. " " is the string containing the character ' '.) Let's look at using a similar function first to get the hang of it. There's a function in a library called equalFilePath :: FilePath -> FilePath -> Bool, which is used to test whether two filenames or folder names represent the same thing. (This solves the problem that on unix, mydir isn't the same as MyDir, but on Windows it is.) Perhaps I want to check a list to see if it's got the file I want:
isMyBestFile :: FilePath -> Bool
isMyBestFile fp = equalFilePath "MyBestFile.txt" fp
but since functions gobble their first argument first, then return a new function to gobble the next, etc, I can write that shorter as
isMyBestFile = equalFilePath "MyBestFile.txt"
This works because equalFilePath "MyBestFile.txt" is itself a function that takes one argument: it's type is FilePath -> Bool. This is partial application, and it's super-useful. Maybe I don't want to bother writing a seperate isMyBestFile function, but want to check whether any of my list has it:
hasMyBestFile :: [FilePath] -> Bool
hasMyBestFile fps = any (equalFilePath "MyBestFile.txt") fps
or just the partially applied version again:
hasMyBestFile = any (equalFilePath "MyBestFile.txt")
Notice how I need to put brackets round equalFilePath "MyBestFile.txt", because if I wrote any equalFilePath "MyBestFile.txt", then filter would try and use just equalFilePath without the "MyBestFile.txt", because functions gobble their first argument first. any :: (a -> Bool) -> [a] -> Bool
Now some functions are infix operators - taking their arguments from before and after, like == or <. In Haskell these are just regular functions, not hard-wired into the compiler (but have precedence and associativity rules specified). What if I was a unix user who never heard of equalFilePath and didn't care about the portability problem it solves, then I would probably want to do
hasMyBestFile = any ("MyBestFile.txt" ==)
and it would work, just the same, because == is a regular function. When you do that with an operator function, it's called an operator section.
It can work at the front or the back:
hasMyBestFile = any (== "MyBestFile.txt")
and you can do it with any operator you like:
hassmalls = any (< 5)
and a handy operator for lists is :. : takes an element on the left and a list on the right, making a new list of the two after each other, so 'Y':"es" gives you "Yes". (Secretly, "Yes" is actually just shorthand for 'Y':'e':'s':[] because : is a constructor/elemental-combiner-of-values, but that's not relevant here.) Using : we can define
space c = c:" "
and we can get rid of the c as usual
space = (:" ")
which hopefully make more sense to you now.
What you want here is an operator section. For that, you'll need to surround the application with parentheses, i.e.
space = (: " ")
which is syntactic sugar for
space = (\x -> x : " ")
(++) won't work here because it expects a string as the first argument, compare:
(:) :: a -> [a] -> [a]
(++) :: [a] -> [a] -> [a]