compare function and parameter in haskell [duplicate]

Question

Possible Duplicate:
Comparing functions in Haskell

I'm learning haskell, wanted to know is something like this possible? All I want is to compare if parameter 'function' is equal to one of the functions a or b. How to do this?

Example code:

a,b :: Integer -> Integer
a x = x+1
b x = x-1

c function parameter = if function == a 
           then ... parameter -- Do a related stuff
           else ... parameter -- Do b related stuff

I'm no Haskell expert, but in order to use the function (==) the parameters have to be part of the Eq typeclass, and I don't think functions can be part of typeclasses. — Vincent Savard
– Vincent Savard, Commented Nov 4, 2012 at 1:10
From this answer: There is not, and will never be, a way to compare two functions for equality. There is a mathematical proof that it is not possible in general. — Miguel
– Miguel, Commented Nov 4, 2012 at 1:12
@VincentSavard: instance Num b => Num (a -> b) where (f + g) x = f x + g x -- ... — Fixnum
– Fixnum, Commented Nov 4, 2012 at 2:01
@Fixnum: Interesting! I'm not quite familiar with the syntax (I can only write really basic programs), but there's a lot of concepts I'll have to look into! — Vincent Savard
– Vincent Savard, Commented Nov 4, 2012 at 2:33

Gabriella Gonzalez · Accepted Answer · 2012-11-04 01:24:22Z

2

The only case I know of where you can compare two functions for equality is if their domain has a finite number of values. For example, if you have two functions of type:

f, g :: Bool -> A

Then they are equal if they are equal for all inputs:

f == g = (f False == g False) && (f True == g True)

However, for the case of Int, comparing them on every possible value of Int is impractical and inefficient. For Integer, it can't be done since Integers are unbounded.

As @Miguel correctly pointed out in his comment, functions with non-finite domains cannot be compared for equality in general.

answered Nov 4, 2012 at 1:24

Gabriella Gonzalez

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