Yet Another Haskell Tutorial/Errata

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Yet Another Haskell Tutorial
Getting Started
Language Basics (Solutions)
Type Basics (Solutions)
IO (Solutions)
Modules (Solutions)
Advanced Language (Solutions)
Advanced Types (Solutions)
Monads (Solutions)
Advanced IO

This page contains a list of errata which have been discovered so far in the PDF version of YAHT (some of these may have already been corrected in the online HTML version - (imported)).

  • Exercise 3.1: "We’ve seen that multiplication binds more tightly than division." - as far as I'm concerned, multiplication and division have the same priority, so they bind equally tightly. Maybe you meant addition vs multiplication or multiplication vs exponentiation? I see it has been fixed in the online version already.
  • There is a bug in the solution to Exercise 7.1 on page 79, under "7.3 Partial Application" (see Advanced Language: Partial Application for the corresponding HTML version (which has already been corrected)), for which the exercise is stated as follows:


Convert the following functions into point-free style, if possible.


func5 f l = foldr (\x y -> f (y,x)) 0 l
  • On page 81, text states

"we create a value, call it x, which has value Red. We then apply this to colorToRGB. We check to see if we can “match” x against Red. This match fails because according to the definition of Eq Color, Red is not equal to Yellow."

Clearly, the premise is wrong, the initial value x should be 'Yellow' in order for the rest of the paragraph to make sense.

  • According to the page 172 of the PDF file, the solution is stated as follows:

func5 = foldr (uncurry $ flip f) 0

However, as pointed out by Michael Mossey on the Haskell-Beginners mailing list, that solution is actually incorrect.

Here is the correct solution:


func5 = flip foldr 0 . flip . curry 

As a sidenote, Daniel Fischer has provided a succinct way of verifying correctness of the solutions, using types, as follows:

A simple way to check for sanity of the results is:


Prelude> :t flip foldr 0 . flip. curry
flip foldr 0 . flip. curry :: (Num c) => ((c, b) -> c) -> [b] -> c

Prelude> :t \f list -> foldr (\x y -> f (y,x)) 0 list
\f list -> foldr (\x y -> f (y,x)) 0 list :: (Num b) =>
                                             ((b, a) -> b) -> [a] -> b

Prelude> :t \f -> foldr (uncurry $ flip f) 0
\f -> foldr (uncurry $ flip f) 0 :: (Num b1) =>
                                    (b -> a -> b1 -> b1) -> [(a, b)] -> b1

So you see that your result has the correct type, while Hal's hasn't.

In the above-mentioned description, the first pair of two lines show that the type of

flip foldr 0 . flip. curry

(i.e., the solution to verify) is

(Num c) => ((c, b) -> c) -> [b] -> c

The second pair of two lines show that the type of

\f list -> foldr (\x y -> f (y,x)) 0 list

(i.e., essentially the expression given in the exercise, which was f l = foldr (\x y -> f (y,x)) 0 l) is also

(Num b) => ((b, a) -> b) -> [a] -> b

(Note that this is the same type as above, only written with renamed variables.)

The third pair of two lines show that the type of

\f -> foldr (uncurry $ flip f) 0

(i.e., the solution given by Hal in the PDF version of the tutorial) is

(Num b1) => (b -> a -> b1 -> b1) -> [(a, b)] -> b1

which is a different type. Hence, the solution given by Hal is incorrect.


(Thanks to both Michael Mossey and Daniel Fischer for their discussion on the Haskell-Beginners mailing list regarding this issue.)

-- Benjamin L. Russell

2. Exercise 4.11:

4.11. Test whether the CPS-style fold mimicks either of foldr and foldl. If not, where is the difference?

It does not specify the "CPS-style fold" function, so the reader should assume that the exercise is about the previously defined cfold:

cfold’ f z [] = z
cfold’ f z (x:xs) = f x z (\y -> cfold’ f y xs)
cfold f z l = cfold’ (\x t g -> f x (g t)) z l

But the solution given in YAHT is actually not about about this function. In fact, this function behaves exactly as foldr. The solution is about very similar function: the only difference is in the order of arguments for the second case of cfold', namely

cfold' f z (x:xs) = f z x (\y -> cfold' f y xs)