I have a recursive function working within the scope of strictly defined interface, so I can't change the function signatures.
The code compiles fine, and even runs fines without error. My problem is that it's a large result set, so it's very hard to test if there is a semantic error.
My primary question is: In a sequence of function calls A to B to A to B to breaking condition, considering the same original Map is passed to all functions until breaking condition, and that some functions only return an Integer, would an insert on a Map in a function that only returns an Integer still be reflected once control is returned to the first function?
primaryFunc :: SuperType -> MyMap -> (Integer, MyMap)
primaryFunc (SubType1 a) mapInstance = do
let returnInt = func1 a mapInstance
(returnInt, mapInstance)
primaryFunc (SubType2 c) mapInstance = do
let returnInt = primaryFunc_nonprefix_SuperType c mapInstance
let returnSuperType = (Const returnInt)
let returnTable = H.insert c returnSuperType mapInstance
(returnInt, returnTable)
primaryFunc (ConstSubType d) mapInstance = do
let returnInt = d
(returnInt, mapInstance)
func1 :: SubType1 -> MyMap -> Integer
func1 oe vt = do
--do stuff with input and map data to get return int
returnInt = primaryFunc
returnInt
func2 :: SubType2 -> MyMap -> Integer
func2 pe vt = do
--do stuff with input and map data to get return int
returnInt = primaryFunc
returnInt
Your question is almost impossibly dense and ambiguous, but it should be possible to answer what you term your "primary" question from the simplest first principles of Haskell:
No Haskell function updates a value (e.g. a map). At most it can return a modified copy of its input.
Outside of the IO monad, no function can have side effects. No function can affect the value of any variable assigned before it was called; all it can do is return a value.
So if you pass a map as a parameter to a function, nothing the function does can alter your existing reference to that value. If you want an updated value, you can only get that from the output of a function to which you have passed the original value as input. New value, new reference.
Because of this, you should have absolute clarity at any depth within your web of functions about which value you are working with. Knowing this, you should be able to answer your own question. Frankly, this is such a fundamental characteristic of Haskell that I am perplexed that you even need to ask.
If a function only returns an integer, then any operations you perform on any values made available to the function can only affect the output - that is, the integer value returned. Nothing done within the function can affect anything else (short of causing the whole program to crash).
So if function A has a reference to a map and it passes this value to function B which returns an int, nothing function B does can affect A's copy of the map. If function B were allowed to secretly alter A's copy of the map, that would be a side effect. Side effects are not allowed.
You need to understand that Haskell does not have variables as you understand them. It has immutable values, references to immutable values and functions (which take inputs and return new outputs). Functions do not have variables which are in scope for other functions which might alter those variables on the fly. That cannot happen.
As an aside, not only does the code you posted show that you do not understand the basics of Haskell syntax, the question you asked shows that you haven't understood the primary characteristics of Haskell as a language. Not only are these fundamentals things which can be understood before having learned any syntax, they are things you need to know to make sense of the syntax.
If you have a deadline, meet it using a tool you do understand. Then go learn Haskell properly.
In addition, you will find that
an insert on a Map in a function that only returns an Integer
is nearly impossible to express. Yes, you can technically do it like in
insert k v map `seq` 42 -- force an insert and throw away the result
but if you think that, for example:
let done = insert k v map in 42
does anything with the map, you're probably wrong.
In no case, however, is the original map altered.
Related
I'm currently working on an assignment. I have a function called gamaTipo that converts the values of a tuple into a data type previously defined by my professor.
The problem is: in order for gamaTipo to work, it needs to receive some preceding element. gamaTipo is defined like this: gamaTipo :: Peca -> (Int,Int) -> Peca where Peca is the data type defined by my professor.
What I need to do is to create a funcion that takes a list of tuples and converts it into Peca data type. The part that im strugling with is taking the preceding element of the list. i.e : let's say we have a list [(1,2),(3,4)] where the first element of the list (1,2) always corresponds to Dirt Ramp (data type defined by professor). I have to create a function convert :: [(Int,Int)] -> [Peca] where in order to calculate the element (3,4) i need to first translate (1,2) into Peca, and use it as the previous element to translate (3,4)
Here's what I've tried so far:
updateTuple :: [(Int,Int)] -> [Peca]
updateTuple [] = []
updateTuple ((x,y):xs) = let previous = Dirt Ramp
in (gamaTipo previous (x,y)): updateTuple xs
Although I get no error messages with this code, the expected output isn't correct. I'm also sorry if it's not easy to understand what I'm asking, English isn't my native tongue and it's hard to express my self. Thank you in advance! :)
If I understand correctly, your program needs to have a basic structure something like this:
updateTuple :: [(Int, Int)] -> [Peca]
updateTuple = go initialValue
where
go prev (xy:xys) =
let next = getNextValue prev xy
in prev : (go next xys)
go prev [] = prev
Basically, what’s happening here is:
updateTuple is defined in terms of a helper function go. (Note that ‘helper function’ isn’t standard terminology, it’s just what I’ve decided to call it).
go has an extra argument, which is used to store the previous value.
The implementation of go can then make use of the previous value.
When go recurses, the recursive call can then pass the newly-calculated value as the new ‘previous value’.
This is a reasonably common pattern in Haskell: if a recursive function requires an extra argument, then a new function (often named go) can be defined which has that extra argument. Then the original function can be defined in terms of go.
Coming from a C# background, I would say that the ref keyword is very useful in certain situations where changes to a method parameter are desired to directly influence the passed value for value types of for setting a parameter to null.
Also, the out keyword can come in handy when returning a multitude of various logically unconnected values.
My question is: is it possible to pass a parameter to a function by reference in Haskell? If not, what is the direct alternative (if any)?
There is no difference between "pass-by-value" and "pass-by-reference" in languages like Haskell and ML, because it's not possible to assign to a variable in these languages. It's not possible to have "changes to a method parameter" in the first place in influence any passed variable.
It depends on context. Without any context, no, you can't (at least not in the way you mean). With context, you may very well be able to do this if you want. In particular, if you're working in IO or ST, you can use IORef or STRef respectively, as well as mutable arrays, vectors, hash tables, weak hash tables (IO only, I believe), etc. A function can take one or more of these and produce an action that (when executed) will modify the contents of those references.
Another sort of context, StateT, gives the illusion of a mutable "state" value implemented purely. You can use a compound state and pass around lenses into it, simulating references for certain purposes.
My question is: is it possible to pass a parameter to a function by reference in Haskell? If not, what is the direct alternative (if any)?
No, values in Haskell are immutable (well, the do notation can create some illusion of mutability, but it all happens inside a function and is an entirely different topic). If you want to change the value, you will have to return the changed value and let the caller deal with it. For instance, see the random number generating function next that returns the value and the updated RNG.
Also, the out keyword can come in handy when returning a multitude of various logically unconnected values.
Consequently, you can't have out either. If you want to return several entirely disconnected values (at which point you should probably think why are disconnected values being returned from a single function), return a tuple.
No, it's not possible, because Haskell variables are immutable, therefore, the creators of Haskell must have reasoned there's no point of passing a reference that cannot be changed.
consider a Haskell variable:
let x = 37
In order to change this, we need to make a temporary variable, and then set the first variable to the temporary variable (with modifications).
let tripleX = x * 3
let x = tripleX
If Haskell had pass by reference, could we do this?
The answer is no.
Suppose we tried:
tripleVar :: Int -> IO()
tripleVar var = do
let times_3 = var * 3
let var = times_3
The problem with this code is the last line; Although we can imagine the variable being passed by reference, the new variable isn't.
In other words, we're introducing a new local variable with the same name;
Take a look again at the last line:
let var = times_3
Haskell doesn't know that we want to "change" a global variable; since we can't reassign it, we are creating a new variable with the same name on the local scope, thus not changing the reference. :-(
tripleVar :: Int -> IO()
tripleVar var = do
let tripleVar = var
let var = tripleVar * 3
return()
main = do
let x = 4
tripleVar x
print x -- 4 :(
Lets suppose we have a function
type Func = Bool -> SophisticatedData
fun1 :: Func
And we'd like to change this function some input:
change :: SophisticatedData -> Func -> Func
change data func = \input -> if input == False then data else func input
Am I right that after several calls of change (endFunc = change data1 $ change data2 $ startFunc) resulting function would call all intermediate ones each time? Am I right that GC wouldn't able to delete unused data? What is the haskell way to cope with this task?
Thanks.
Well let's start by cleaning up change to be a bit more legible
change sd f input = if input then func input else sd
So when we compose these
change d1 $ change d2 $ change d3
GHC starts by storing a thunk for each of them. Remember that $ is a function to so the whole change d* thing is going to be a thunk at first. Thunks are relatively cheap and if you're not creating 10k or so of them at once you'll be just fine :) so no worries there.
Now the question is, what happens when you start evaluating, the answer is, it'll still not evaluate the complex data, so it's still quite memory efficient, and it only needs to force input to determine which branch it's taking. Because of this, you should never actually fully evaluate SophisticatedData until after choose has run and returned a one to you, then it will be evaluated as need if you use it.
Further more, at each step, GHC can garbage collect the unneeded thunks since they can't be referenced anymore.
In conclusion, you should be just fine. Trust in the laziness
You are correct: if foo is a chain of O(n) calls to change, there will be O(n) overhead on every call to foo. The way to deal with this is to memoize foo:
memoize :: Func -> Func
memoize f = \x -> if x then fTrue else fFalse where
fTrue = f True
fFalse = f False
I'm looking to call functions dynamically based on the contents found in an association list.
Here is an example in semi-pseudo-code. listOfFunctions would be passed to callFunctions.
listOfFunctions = [('function one', 'value one')
, ('function two', 'value two')
, ('function three', 'value three')]
callFunctions x = loop through functions
if entry found
then call function with value
else do nothing
The crux of the question is not looping through the list, rather, it's how to call a function once I have it's name?
Consider this use case for further clarification. You open the command prompt and are presented with the following menu.
1: Write new vHost file
2: Exit
You write the new vHost file and are not presented with a new menu
1: Enter new directive
2: Write file
3: Exit
You enter some new directives for the vHost and are now ready to write the file.
The program isn't going to blindly write each and every directive it can, rather, it will only write the ones that you supplied. This is where the association list comes in. Writing a giant if/then/else or case statement is madness. It would be much more elegant to loop through the list, look for which directives were added and call the functions to write them accordingly.
Hence, loop, find a function name, call said function with supplied value.
Thanks to anyone who can help out with this.
Edit:
Here is the solution that I've come up with (constructive critiques are always welcome).
I exported the functions which write the directives in an association list as every answer provided said that just including the function is the way to go.
funcMap = [("writeServerName", writeServerName)
,("writeServeralias", writeServerAlias)
,("writeDocRoot", writeDocRoot)
,("writeLogLevel", writeErrorLog)
,("writeErrorPipe", writeErrorPipe)
,("writeVhostOpen", writeVhostOpen)]
In the file which actually writes the hosts, that file is imported.
I have an association list called hostInfo to simulate some dummy value that would be gathered from an end-user and a function called runFunction which uses the technique supplied by edalorzo to filter through both the lists. By matching on the keys of both lists I ensure that the right function is called with the right value.
import Vhost.Directive
hostInfo = [("writeVhostOpen", "localhost:80")
,("writeServerName", "norics.com")]
runFunctions = [f val | (mapKey, f) <- funcMap, (key, val) <- hostInfo, mapKey == key]
You can simply include the function in the list directly; functions are values, so you can reference them by name in a list. Once you've got them out of the list, applying them is just as simple as func value. There's no need to involve their names at all.
Since I am farily new to Haskell I will risk that you consider my suggestion very naive, but anyways here it goes:
let funcs = [("sum", (+3),1),("product", (*3),2),("square", (^2),4)]
[f x | (name, f, x) <- funcs, name == "sum"]
I think it satisfies the requirements of the question, but perhaps what you intend is more sofisticated than what I can see with my yet limitted knowledge of Haskell.
It might be a bit of an overkill (I agree with ehird's reasoning) but you can evaluate a string with Haskell code by using the eval function in System.Eval.Haskell.
EDIT
As pointed out in the comments, hint is a better option for evaluating strings with Haskell expressions. Quoting the page:
This library defines an Interpreter monad. It allows to load Haskell modules, browse them, type-check and evaluate strings with Haskell expressions and even coerce them into values. The library is thread-safe and type-safe (even the coercion of expressions to values). It is, esentially, a huge subset of the GHC API wrapped in a simpler API. Works with GHC 6.10.x and 6.8.x
First we define our list of functions. This could be built using more machinery, but for the sake of example I just make one explicit list:
listOfFunctions :: [(Int, IO ())]
listOfFunctions = [(0, print "HI") -- notice the anonymous function
,(1, someNamedFunction) -- and something more traditional here
]
someNamedFunction = getChar >>= \x -> print x >> print x
Then we can select from this list however we want and execute the function:
executeFunctionWithVal :: Int -> IO ()
executeFunctionWithVal v = fromMaybe (return ()) (lookup v listOfFunctions)
and it works (if you import Data.Maybe):
Ok, modules loaded: Main.
> executeFunctionWithVal 0
"HI"
> executeFunctionWithVal 01
a'a'
'a'
Don't store the functions as strings, or rather, try storing the actual functions and then tagging them with a string. That way you can just call the function directly. Functions are first class values, so you can call the function using whatever name you assign it to.
I'm aware of partial updates for records like :
data A a b = A { a :: a, b :: b }
x = A { a=1,b=2 :: Int }
y = x { b = toRational (a x) + 4.5 }
Are there any tricks for doing only partial initialization, creating a subrecord type, or doing (de)serialization on subrecord?
In particular, I found that the first of these lines works but the second does not :
read "A {a=1,b=()}" :: A Int ()
read "A {a=1}" :: A Int ()
You could always massage such input using a regular expression, but I'm curious what Haskell-like options exist.
Partial initialisation works fine: A {a=1} is a valid expression of type A Int (); the Read instance just doesn't bother parsing anything the Show instance doesn't output. The b field is initialised to error "...", where the string contains file/line information to help with debugging.
You generally shouldn't be using Read for any real-world parsing situations; it's there for toy programs that have really simple serialisation needs and debugging.
I'm not sure what you mean by "subrecord", but if you want serialisation/deserialisation that can cope with "upgrades" to the record format to contain more information while still being able to process old (now "partial") serialisations, then the safecopy library does just that.
You cannot leave some value in Haskell "uninitialized" (it would not be possible to "initialize" it later anyway, since Haskell is pure). If you want to provide "default" values for the fields, then you can make some "default" value for your record type, and then do a partial update on that default value, setting only the fields you care about. I don't know how you would implement read for this in a simple way, however.