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 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 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 http://www.haskell.org/mailman/listinfo/haskell-cafe

-- Ozgur Akgun

-- Mujtaba Ali Alboori

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 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 > 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 mailto:mujtaba.boori@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

Great job Stephen. Thank for explaining . I got it to work. On Tue, Apr 20, 2010 at 9:21 AM, Stephen Tetley wrote:

Hi

If you are working with characteristic functions (Point -> Bool or Point -> Colour...) the common way to do this is to manufacture a Num instance for functions. This gives you syntax overloading of the (+, -, *) operators. Similarly you might want to overload (or have to overload) Floating, Fractional...

Examples using this technique are Jerzy Karczmarczuk's Clastic, Conal Elliott's Vertigo, Tangible Values, Pan etc.

To overload Num you have to define Show and Eq instances for functions as well. Something along the lines of this is adequate:

type CF = (Double,Double) -> Bool

instance Show CF where show _ = "<function>"

instance Eq CF where (==) _ _ = error "No Eq on Characteristic functions"

instance Num CF where f + g = \pt -> f pt + g pt -- ... negate f = \(x,y) -> f (negate x, negate y)

-- ... rest follows this pattern, Floating, Fractional similar

If you characteristic function is Point -> Bool then you also need a Num instance for Bool.

All that said, I think your formulation of func above is slightly wrong to fit this style. Its forming a function (-> Point) "to point" rather than a characteristic function Point -> Bool.

Best wishes

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

-- Mujtaba Ali Alboori