Using composition can be tricky with more than one arg. I just want to be sure you're not really looking for something like:

func :: (a -> Bool) -> (b -> Bool) -> (a -> b -> Bool)

keeping with your given type I think you're looking for something like:

func f1 f2 x = (f1 x) || (f2 x)

I'm sure there is a nice way to do this with function composition but I still find composition less intuitive than explicit args in cases like this. On Sun, Apr 18, 2010 at 1:00 PM, Mujtaba Boori wrote:

Thanks for helping me but I have another problem (sorry for asking) . I tried to figure it out . how about if I want to compare two kind with (&&) (||)  for func :: (a -> Bool) -> (a -> Bool) -> (a -> Bool)

I tried some thing like func = ((||) .) This is the annoying part about Haskell . I can not understand composition .

On Sun, Apr 18, 2010 at 4:35 PM, Mujtaba Boori wrote:

...

Hello I am kinda newbie in Haskell you can help help me with some programming I am trying to make function like for example func :: (a -> Bool) -> (a -> Bool) this function make calculation  and return bool . I want to be able to make bool True when It is False and False when it is True while returning the a. Thank you -- Mujtaba Ali Alboori

-- Mujtaba Ali Alboori

_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

-- keithsheppard.name

Thanks a lot guys you were really helpful func f1 f2 x = (f1 x) || (f2 x) is working for me On Sun, Apr 18, 2010 at 6:27 PM, Thomas Davie wrote:

To do this, you need not just fmap (composition), but also ap, or the combined form, liftA2:

func = liftA2 (||)

Bob

On 18 Apr 2010, at 18:21, Keith Sheppard wrote:

...

Using composition can be tricky with more than one arg. I just want to be sure you're not really looking for something like:

...

func :: (a -> Bool) -> (b -> Bool) -> (a -> b -> Bool)

keeping with your given type I think you're looking for something like:

...

func f1 f2 x = (f1 x) || (f2 x)

I'm sure there is a nice way to do this with function composition but I still find composition less intuitive than explicit args in cases like this.

On Sun, Apr 18, 2010 at 1:00 PM, Mujtaba Boori wrote:

...

Thanks for helping me but I have another problem (sorry for asking) . I tried to figure it out . how about if I want to compare two kind with (&&) (||) for func :: (a -> Bool) -> (a -> Bool) -> (a -> Bool)

I tried some thing like func = ((||) .) This is the annoying part about Haskell . I can not understand composition .

On Sun, Apr 18, 2010 at 4:35 PM, Mujtaba Boori

...

wrote:

...

Hello I am kinda newbie in Haskell you can help help me with some programming I am trying to make function like for example func :: (a -> Bool) -> (a -> Bool) this function make calculation and return bool . I want to be able to make bool True when It is False and False when it is True while

returning

...

the a. Thank you -- Mujtaba Ali Alboori

-- Mujtaba Ali Alboori

_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

-- keithsheppard.name _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

-- Mujtaba Ali Alboori

On Sun, Apr 18, 2010 at 10:25 AM, Sean Leather wrote:

...

This is the annoying part about Haskell . I can not understand composition .

One of the ways of understanding composition (and many other functions in Haskell) is by trying to understand its type. Here it is shown by looking at the type in the interpreter GHCi.

*Main> :t (.) (.) :: (b -> c) -> (a -> b) -> a -> c

It's a function that takes three arguments, the first two of which are functions, and the third is something else.

As you mention below the function arrow (->), is right associative. For this reason I like to think of the above function as binary, instead of a function of 3 arguments. Making the associativity explicit we can rewrite it: (.) :: (b -> c) -> (a -> b) -> (a -> c) Now we can see that it is a binary function on functions. You're mileage may vary, but I find this form more intuitive. Jason