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)).

- 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:

Exercises |
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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:

**Example:**

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.

*Note*

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

-- Benjamin L. Russell

YAHT has never been comprehensively proofread, so this is only a partial list. |

This is still a work in progress. Feel free to add to it as needed! |

2. Exercise 4.11:

Exercises |
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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)