Thanks Dan. Great help
but my problem has not solved yet
This doesn't work for type ((Float -> Float)->Bool)
to make it easier ignore the rotation and suppose I want just multiplay with
whatever (x ,y) and return the result to this type ((Float -> Float)->Bool)
note this type is shorten and replace by t
Type Point = (Float, Float)
Type Bitmap = Point -> Bool
so the function type actually
func :: Bitmap -> Float -> Bitmap
I want to take Bitmap do some calculation on Bitmap the return it
as Bitmap.
GHCi response for Dan method is this
Couldn't match expected type `Bitmap'
against inferred type `(a, b)'
so it is missing a Bool.
hopefully it is clear .
On Mon, Apr 19, 2010 at 7:02 PM, Dan Weston <weston...@imageworks.com>wrote:
> First of all, your function
>
> func (x,y) s dg =((x*(cos dg) - y*(sin dg)),(x*(sin dg) - y*(cos dg)))
> does NOT work for type (Float -> Float), unless you mean that that is the
> type of the unused parameter s. Also, your desired type ((Float -> Float) ->
> Bool) itself looks suspicious. It must accept any function (without
> something to apply it to) and arbitrarily return True or False. How will you
> decide which? I suspect you need another parameter for this function.
>
> Second, on the off chance you are trying to calculate the position on a
> circle scaled then rotated an angle dg from (x,y), that new position is
>
> f (x,y) s dg = (s*(x*(cos dg) - y*(sin dg)),s*(x*(sin dg) + y*(cos dg)))
>
> in which case you are missing the s and the last minus sign in your formula
> should be a plus sign.
> If so, this can be evaluated with greater clarity (and probably accuracy)
> in polar coordinates:
>
> g (x,y) s dg = (r * cos a, r * sin a)
> where r = s * sqrt (x^2 + y^2)
> a = atan2 y x + dg
>
> Third, if you did not need the scale, I would use an underscore to make
> that clear:
>
> h (x,y) _ dg = (r * cos a, r * sin a)
> where r = sqrt (x^2 + y^2)
> a = atan2 y x + dg
>
> That's all the observations I can make unless you describe the problem more
> clearly. Sorry.
>
> Dan
>
> Mujtaba Boori wrote:
>
>> sorry
>> ok I am trying to make these calculation
>> func (x,y) s dg =((x*(cos dg) - y*(sin dg)),(x*(sin dg) - y*(cos dg)))
>>
>> This work for type (Float -> Float)
>>
>> but how can make it work with ((Float -> Float) -> Bool)
>>
>> because my main function that I want use with. it takes (Float,Float)
>> ->Bool) I need to return the same type ((Float,Float) ->Bool) so it could
>> be used with other function.
>>
>> On Mon, Apr 19, 2010 at 5:54 PM, Ozgur Akgun <ozgurak...@gmail.com<mailto:
>> ozgurak...@gmail.com>> wrote:
>>
>> Can you at least give an example of how you intend to use this "func"?
>> Since you do not describe it's behaviour, it is very hard to make a
>> useful
>> comment (at least for me)
>>
>> Best,
>>
>> On 19 April 2010 16:54, Mujtaba Boori <mujtaba.bo...@gmail.com
>> <mailto:mujtaba.bo...@gmail.com>> wrote:
>> >
>> > Hello
>> > I am sorry for the silly question.
>> >
>> > I have a function as the following
>> > func:: ((Float,Float) ->Bool) -> Float -> ((Float,Float) -> Bool)
>> > I am trying to make calculation in this type ((Float,Float)
>> ->Bool) with Float and then pass the information to ((Float,Float)
>> -> Bool)
>> >
>> > Thank again appreciated.
>> > _______________________________________________
>> > Haskell-Cafe mailing list
>> > Haskell-Cafe@haskell.org <mailto:Haskell-Cafe@haskell.org>
>>
>> > http://www.haskell.org/mailman/listinfo/haskell-cafe
>> >
>>
>>
>>
>> --
>> Ozgur Akgun
>>
>>
>>
>>
>> --
>> Mujtaba Ali Alboori
>>
>>
--
Mujtaba Ali Alboori
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