[Haskell-cafe] Newbie: Applying Unknown Number Arguments to A Partial Function

Arthur Baars arthurb at cs.uu.nl
Fri May 19 05:24:42 EDT 2006

Chris, the subject states clearly that Aditya is a Newbie, and is most 
likely just trying to define the function "map". So I think pointing to 
a bunch of advanced type magic tricks is not really helpful.

Aditya, you say you want the function applyArgument to take a function 
and a list and apply the function to all elements of that list. The 
result of "applyArgument" is a list, with the results of applying the 
function to each element.

So you want the type of "applyArgument" to be:
applyArgument :: (a->b) -> [a] -> [b]

There are a numbers of errors in your code.
applyArgument f (arg) = f arg
The variable "f" is the first parameter of "applyArgument", and has 
type (a -> b)
The variable "arg" is the second parameter of "applyArgument", and has 
type [a] .
You try to apply "f" to a list which is not ok.
Most likely you meant "[arg]", which is a singleton list, instead of 
"(arg)", which is the same as just "arg"
Furthermore "applyArgument" returns a list of result. The function "f" 
only yields a "b", instead of a list "[b]"

The following does work:
applyArgument f [arg] = [f arg]

In your second line:
applyArgument f (arg:args) = applyArgument (f arg) args
you use a list pattern "(arg:args)", which is good. The variable "arg" 
is the head of the list, and the variable "args" is the tail of the 
list. So "arg" has type "a", and "args" has type "[a]" . The 
application  "(f arg)" has type "b". Because the function 
"applyArgument" expects a function as first argument and not a "b", so 
this is wrong. I guess you can find out by your self how to fix this. 
There are a number of very good tutorials for beginners on 
http://haskell.org .

Finally your function won't work for empty lists, it is only defined 
for singleton lists and longer lists. You can fix this by replacing the 
definition for singleton lists with a definition for the empty list. 
The pattern for empty list is simply "[]".

Good luck,


On 19-mei-06, at 10:10, Chris Kuklewicz wrote:

> Aditya Siram wrote:
>> I am trying to write a function 'applyArguments' which takes a 
>> function
>> and a list and recursively uses element each in the list as an 
>> argument
>> to the function. I want to do this for any function taking any number 
>> of
>> arguments.
>> applyArgument f (arg) = f arg
>> applyArgument f (arg:args) = applyArgument (f arg) args
>> This has failed in Hugs, so my question is: Can I conceptually do 
>> this?
>> If so, what is the type signature of this function?
>> Deech
> You can't do that, but there are other tricks that do work:
> http://okmij.org/ftp/Haskell/types.html
> which describes "Functions with the variable number of (variously 
> typed)
> arguments" and "Genuine keyword arguments"
> -- 
> Chris
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