Which parts would like explanations for? All of them? On Sun, Sep 27, 2009 at 12:22 PM, informationen wrote:

Hi,

i am trying to understand the Haskell type system. I thought i understood it quite well until i encountered the three following exercises. As you can see, i have the answers already. But i don't understand, why they are correct.

Could anybody tell me a good place where i could learn how to answers these kind of questions correctly or could give me some explanations, why these answers are correct?

Any help is highly appreciated.

Kind regards

Chris Two functions f and g with the following signatures are given: f :: a -> a g :: b -> c -> b

A) Give the type of the following expressions:

1) g [False] True :: 2) g [] True :: 3) g (f True) :: 4) g f :: Answers:

1 [Boolean] 2) [a] 3) c -> Bool 4) c -> (a -> a)1) g f 1 B) Which of the following statements is correct?

2) g (f 1) is type correct 3) g . (f 1) is type correct 4) g . f 1 is type correct 5) (g . f) 1 is type correct 6) none of the expressions is correct

Answers: 1,2 and 5 are correct.

C) A function h is given as: h p x = p (f x). Which of the following statements is correct.

1) h :: a -> b -> a -> b 2) h :: (a -> a) -> a -> a 3) h :: (a -> b) -> a -> b 4) h is equivalent to h' with h' p = p . f 5) h is equivalent to h' with h' p = p f 5) h is equivalent to h' with h' p x = p f x

Answers: (I am not sure, if i remember correctly, but 3) and 4) should be correct.)

_______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners

I'll try my hand at these, anybody feel free to point out all my deliberate mistakes. On 27 Sep 2009, at 11:22, informationen wrote:

Two functions f and g with the following signatures are given: f :: a -> a g :: b -> c -> b

A) Give the type of the following expressions:

1) g [False] True :: 2) g [] True

These two are simply that the result of the function g is the same as it's first argument. The first takes a list of Boolean, the second a list of an unknown type. Therefore the results are the same types [Boolean] and [a] respectively.

:: 3) g (f True) ::

The result of f is the same type as it's argument, in this case Boolean. g's first and only argument which is the type of the result of f -- Boolean. As we haven't specified a second argument, the returned type is a function which takes an unknown type and returns a boolean. (c -> Bool)

4) g f ::

Similar to the last case except now the only argument supplied is a function of type (a -> a). (c -> (a -> a))

Answers:

B) Which of the following statements is correct?

First two are easy, and the same.

1) g f 1 c -> Int 2) g (f 1) c -> Int

Remember that the type of (.) is (b -> c) -> (a -> b) -> a -> c. That is, it takes two functions where they both take an argument, and the type of the first functions argument is the same as the seconds result.

3) g . (f 1)

Here (f 1) doesn't take an argument, it's already supplied. That makes it's type Int, not (a -> b). That's not right.

4) g . f 1

The same, the brackets where redundant in the previous expression.

5) (g . f) 1

Here we're composing before giving the arguments. So, remembering that f :: a -> a and g :: b -> c -> b, we can start filling in the types: (b -> c) -> (a -> b) -> a -> c -- To make g fit, c must become (c -> b) (b -> (c -> b)) -> (a -> b) -> a -> (c -> b) -- f returns same type as input, so b == a (a -> (c -> a)) -> (a -> a) -> a -> (c -> a) -- Now we've supplied f and g we can just worry about the result part a -> (c -> a) -- This is the type of (g . f), but we've got an Int argument "1" Int -> (c -> Int) -- So the type of the expression is (c->Int) Nothing wrong with that. Type safe. It might be easier to just start plugging types in than following that explanation, but it does work.

C) A function h is given as: h p x = p (f x). Which of the following statements is correct.

p is obviously a function of one (or more) arguments. h takes two arguments, and returns the result of p, so h :: (a -> b) -> c -> b because we know f :: a -> a, we can say a and c are the same type h :: (a -> b) -> a -> b That's number 3. (g . f) x is the same as g (f x). Therefore we can re-write h h' p x = (p . f) x -- or h' p = (p . f) -- the argument is implicit That's 4

3) h :: (a -> b) -> a -> b 4) h is equivalent to h' with h' p = p . f

I have found this online book to be fairly easy to understand http://book.realworldhaskell.org/read/types-and-functions.html http://book.realworldhaskell.org/read/types-and-functions.htmlTake a look at the section called "The type of a function with more than one argument" and also follow the link to "Partial function application and currying" If you decide to use this book, I would recommend reading the entire thing. I tried to skip around at first, but got lost pretty quickly. The author does a good job of adding complexity step by step, so it might be best to start from the beginning. On Sun, Sep 27, 2009 at 3:22 AM, informationen wrote:

Hi,

i am trying to understand the Haskell type system. I thought i understood it quite well until i encountered the three following exercises. As you can see, i have the answers already. But i don't understand, why they are correct.

Could anybody tell me a good place where i could learn how to answers these kind of questions correctly or could give me some explanations, why these answers are correct?

Any help is highly appreciated.

Kind regards

Chris Two functions f and g with the following signatures are given: f :: a -> a g :: b -> c -> b

A) Give the type of the following expressions:

1) g [False] True :: 2) g [] True :: 3) g (f True) :: 4) g f :: Answers:

1 [Boolean] 2) [a] 3) c -> Bool 4) c -> (a -> a)1) g f 1 B) Which of the following statements is correct?

2) g (f 1) is type correct 3) g . (f 1) is type correct 4) g . f 1 is type correct 5) (g . f) 1 is type correct 6) none of the expressions is correct

Answers: 1,2 and 5 are correct.

C) A function h is given as: h p x = p (f x). Which of the following statements is correct.

1) h :: a -> b -> a -> b 2) h :: (a -> a) -> a -> a 3) h :: (a -> b) -> a -> b 4) h is equivalent to h' with h' p = p . f 5) h is equivalent to h' with h' p = p f 5) h is equivalent to h' with h' p x = p f x

Answers: (I am not sure, if i remember correctly, but 3) and 4) should be correct.)

_______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners