I see the point with :t (.) that max.(+1) 2 2 is the same as max. 3 2 Which is not what I want. But I have no idea what the type signature of this expression mean now Prelude> :t max. 3 2 max. 3 2 :: (Ord b, Num a1, Num (a1 -> a -> b)) => a -> b -> b Any idea? Daryoush On Thu, Nov 15, 2012 at 11:19 PM, Ramana Kumar wrote:

Hi Daryoush,

I recommend you try these experiments first, and then reply back if you're still confused.

:t max

:t (+1)

:t max . (+1)

:t (+1) 2

:t (.)

On Fri, Nov 16, 2012 at 7:10 AM, Daryoush Mehrtash wrote:

...

I am having hard time understanding how removing the outer parenthesis in

(max.(+1)) 2 2

to

max.(+1) 2 2

changes the meaning of expression.

My expectation was that "max.(+1) takes two numbers and returns the max as defined in the type:

:t max.(+1) max.(+1) :: (Ord b, Num b) => b -> b -> b

With parenthesis it does what I expect it to:

Prelude> :t (max.(+1)) 2 2 (max.(+1)) 2 2 :: (Ord b, Num b) => b Prelude> (max.(+1)) 2 2 3

But if I remove the parenthesis I get a beast that I have no idea what its type signature mean any more

Prelude> :t (max.(+1)) 2 2 (max.(+1)) 2 2 :: (Ord b, Num b) => b Prelude> :t max.(+1) 2 2 max.(+1) 2 2 :: (Ord b, Num a1, Num (a1 -> a -> b)) => a -> b -> b

How did removal of parenthesis changed the meaning? How do you interpret the type: "(Ord b, Num a1, Num (a1 -> a -> b)) => a -> b -> b"?

Thanks

-- Daryoush

Weblog: http://onfp.blogspot.com/

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

-- Daryoush Weblog: http://onfp.blogspot.com/

Yes, same problem, and again I have no idea how to read the type Prelude> :t max. 3 2 max. 3 2 :: (Ord b, Num a1, Num (a1 -> a -> b)) => a -> b -> b What does the type mean in plain english? Daryoush On Thu, Nov 15, 2012 at 11:56 PM, Ramana Kumar wrote:

Another experiment may be revealing:

:t 3 2

On Fri, Nov 16, 2012 at 7:46 AM, Daryoush Mehrtash wrote:

...

I see the point with :t (.) that

max.(+1) 2 2

is the same as

max. 3 2

Which is not what I want.

But I have no idea what the type signature of this expression mean now

Prelude> :t max. 3 2 max. 3 2 :: (Ord b, Num a1, Num (a1 -> a -> b)) => a -> b -> b

Any idea?

Daryoush

On Thu, Nov 15, 2012 at 11:19 PM, Ramana Kumar

...

wrote:

...

Hi Daryoush,

I recommend you try these experiments first, and then reply back if you're still confused.

:t max

:t (+1)

:t max . (+1)

:t (+1) 2

:t (.)

On Fri, Nov 16, 2012 at 7:10 AM, Daryoush Mehrtash wrote:

...

I am having hard time understanding how removing the outer parenthesis in

(max.(+1)) 2 2

to

max.(+1) 2 2

changes the meaning of expression.

My expectation was that "max.(+1) takes two numbers and returns the max as defined in the type:

:t max.(+1) max.(+1) :: (Ord b, Num b) => b -> b -> b

With parenthesis it does what I expect it to:

Prelude> :t (max.(+1)) 2 2 (max.(+1)) 2 2 :: (Ord b, Num b) => b Prelude> (max.(+1)) 2 2 3

But if I remove the parenthesis I get a beast that I have no idea what its type signature mean any more

Prelude> :t (max.(+1)) 2 2 (max.(+1)) 2 2 :: (Ord b, Num b) => b Prelude> :t max.(+1) 2 2 max.(+1) 2 2 :: (Ord b, Num a1, Num (a1 -> a -> b)) => a -> b -> b

How did removal of parenthesis changed the meaning? How do you interpret the type: "(Ord b, Num a1, Num (a1 -> a -> b)) => a -> b -> b"?

Thanks

-- Daryoush

Weblog: http://onfp.blogspot.com/

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

-- Daryoush

Weblog: http://onfp.blogspot.com/

-- Daryoush Weblog: http://onfp.blogspot.com/

On Fri, Nov 16, 2012 at 8:00 AM, Daryoush Mehrtash wrote:

Sorry wrong paste

Prelude> :t 3 2 3 2 :: (Num a, Num (a -> t)) => t

It means that 2 has type a (which must be in the Num class), 3 has type (a -> t) (which also must be in the Num class), and the type of (3 2) is therefore t. When you apply one expression to another, the first must have function type, and the second must have the type of the domain of the function, hence the a -> t and a. Furthermore, the numeric literals are always interpreted as being in the Num class. I haven't been able to find any good introduction to numeric literals and the Num class with quick search, so if anyone else on list can point us to one (if it exists) that would be great.

What does the type mean in plain english?

Daryoush

On Fri, Nov 16, 2012 at 12:00 AM, Daryoush Mehrtash wrote:

...

Yes, same problem, and again I have no idea how to read the type

Prelude> :t max. 3 2 max. 3 2 :: (Ord b, Num a1, Num (a1 -> a -> b)) => a -> b -> b

What does the type mean in plain english?

Daryoush

On Thu, Nov 15, 2012 at 11:56 PM, Ramana Kumar

...

wrote:

...

Another experiment may be revealing:

:t 3 2

On Fri, Nov 16, 2012 at 7:46 AM, Daryoush Mehrtash wrote:

...

I see the point with :t (.) that

max.(+1) 2 2

is the same as

max. 3 2

Which is not what I want.

But I have no idea what the type signature of this expression mean now

Prelude> :t max. 3 2 max. 3 2 :: (Ord b, Num a1, Num (a1 -> a -> b)) => a -> b -> b

Any idea?

Daryoush

On Thu, Nov 15, 2012 at 11:19 PM, Ramana Kumar < Ramana.Kumar@cl.cam.ac.uk> wrote:

...

Hi Daryoush,

I recommend you try these experiments first, and then reply back if you're still confused.

:t max

:t (+1)

:t max . (+1)

:t (+1) 2

:t (.)

On Fri, Nov 16, 2012 at 7:10 AM, Daryoush Mehrtash < dmehrtash@gmail.com> wrote:

...

I am having hard time understanding how removing the outer parenthesis in

(max.(+1)) 2 2

to

max.(+1) 2 2

changes the meaning of expression.

My expectation was that "max.(+1) takes two numbers and returns the max as defined in the type:

:t max.(+1) max.(+1) :: (Ord b, Num b) => b -> b -> b

With parenthesis it does what I expect it to:

Prelude> :t (max.(+1)) 2 2 (max.(+1)) 2 2 :: (Ord b, Num b) => b Prelude> (max.(+1)) 2 2 3

But if I remove the parenthesis I get a beast that I have no idea what its type signature mean any more

Prelude> :t (max.(+1)) 2 2 (max.(+1)) 2 2 :: (Ord b, Num b) => b Prelude> :t max.(+1) 2 2 max.(+1) 2 2 :: (Ord b, Num a1, Num (a1 -> a -> b)) => a -> b -> b

How did removal of parenthesis changed the meaning? How do you interpret the type: "(Ord b, Num a1, Num (a1 -> a -> b)) => a -> b -> b"?

Thanks

-- Daryoush

Weblog: http://onfp.blogspot.com/

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

-- Daryoush

Weblog: http://onfp.blogspot.com/

-- Daryoush

Weblog: http://onfp.blogspot.com/

-- Daryoush

Weblog: http://onfp.blogspot.com/

Let's see tthis: Prelude> :t 3 "a" 3 "a" :: (Num ([Char] -> t)) => t No complaint from GHC; but now see this: Prelude> :t "a" 3 <interactive>:1:0: Couldn't match expected type `t1 -> t' against inferred type `[Char]' In the expression: "a" 3 Why does it not fail for (:t 3 "a") but does fail for (:t "a" 3)? In the second case, why the GHC doesn't give something like? ([Char] (a -> t), Num a) => t In other words, how the first one does make "sense" (however strange the "sense" may seem) to GHC while the second one doesn't? Now let's consider the evaluation scenario: Prelude> 3 2 <interactive>:1:0: No instance for (Num (t -> t1)) arising from the literal `3' at <interactive>:1:0-2 Possible fix: add an instance declaration for (Num (t -> t1)) In the expression: 3 2 In the definition of `it': it = 3 2 Note, how the GHC suggests us this "strange constraint" as a *possible fix*BUT see this: Prelude> "a" 2 <interactive>:1:0: Couldn't match expected type `t1 -> t' against inferred type `[Char]' In the expression: "a" 2 In the definition of `it': it = "a" 2 In this case the GHC doesn't even allow us to add any "possible fix". -Damodar On Fri, Nov 16, 2012 at 2:27 PM, Sean Leather wrote:

Hi Daryoush,

Prelude> :t 3 2

...

3 2 :: (Num a, Num (a -> t)) => t

What does the type mean in plain english?

It's important to remember that numeric literals are polymorphic. That is, 3 :: Num a => a. They do not have monomorphic types such as Int or Integer.

In the above, GHCi is inferring the principal type of 3 applied to 2. Since 3 is in the position of function application, it should have a function type, e.g. a -> t. And 2 is the argument to 3, so it has the type 'a'. But there must be Num constraints on these types, and that's what the context (Num a, Num (a -> t)) is telling you. Num (a -> t) is a strange constraint but so is applying 3 to 2. Whenever you see a Num constraint on a function type, it probably means you're doing something wrong.

You might find the (brief) description of typing numeric literals in the language definition helpful:

http://www.haskell.org/onlinereport/haskell2010/haskellch6.html#x13-1360006....

Also, in response to your initial query...

I am having hard time understanding how removing the outer parenthesis in

...

(max.(+1)) 2 2 to max.(+1) 2 2

Keep in mind the precedence of the function composition operator (.) here:

Prelude> :i (.) (.) :: (b -> c) -> (a -> b) -> a -> c -- Defined in GHC.Base infixr 9 .

Function application is infixl 10, so even though max . (+1) :: (Num b, Ord b) => b -> b -> b, when you do max . (+1) 2, the application of (+1) to 2 binds more tightly than (.).

A common idiom for removing parentheses with a sequence of compositions is to use ($):

Prelude> :i ($) ($) :: (a -> b) -> a -> b -- Defined in GHC.Base infixr 0 $

Note that it has the lowest possible precedence. So, you often see the composition of functions as f . g . h $ arg. But this doesn't work well for (curried) functions with 2 arguments. So, in your case, I'd guess the parentheses is simplest idiomatic solution. Though, I think I'd prefer max (succ 2) 2.

Regards, Sean

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

-- Thanks and regards, -Damodar Kulkarni

On 12-11-17 02:19 AM, damodar kulkarni wrote:

Let's see tthis: Prelude> :t 3 "a" 3 "a" :: (Num ([Char] -> t)) => t

No complaint from GHC; but now see this:

Prelude> :t "a" 3

<interactive>:1:0: Couldn't match expected type `t1 -> t' against inferred type `[Char]' In the expression: "a" 3

Why does it not fail for (:t 3 "a") but does fail for (:t "a" 3)?

3 is polymorphic, "a" is monomorphic (exactly [Char]). To make "a" polymorphic, turn on OverloadedStrings: :set -XOverloadedStrings :type "a" "a" :: Data.String.IsString a => a :type "a" 3 "a" 3 :: (Num a, Data.String.IsString (a -> t)) => t Success! This is clearly depravity.