Haskell：检查两个列表是否相等

Question

I want to check if two lists A and B are equal, ie, a1 == b1, a2 == b2 ,...

I have a working solution:

all (\x->x) zipWith $ (==) A B

Another idea is to do it recursively: a:as, b:bs ; check if a1==b1 and call the function with the remaining lists as and bs . But isn't there an easier and more readable way to do this?

Answer 1

You can just use == on them directly.

> [1, 2, 3] == [1, 2, 3]
True
> [1, 2, 3] == [1, 2]
False

This is because == is part of the Eq type class, and there is an Eq instance for lists which looks something like this:

instance Eq a => Eq [a]

This means that lists instantiate Eq as long as the element type also instantiates Eq , which is the case for all types defined in the standard Prelude except functions and IO actions.

Answer 2

First, hammar's answer is correct, so accept his answer please. (Edit: Which you have done, thank you.)

listA == listB

(I'm going to nitpick on small details in your question, mostly for the benefit of future beginners who find this page on Google.)

Second, A and B aren't lists: they start with upper case letters, so they cannot be variables. I'm going to call them listA and listB instead.

Third, there is a typo in your working solution: the $ should be before the zipWith , not after. The way it appears in your question results in a compile error. I think you meant this:

all (\x->x) $ zipWith (==) listA listB

Fourth, (\\x->x) is better known as the function id .

all id $ zipWith (==) listA listB

Fifth, as Matvey points out, all id is the same as and .

and $ zipWith (==) listA listB

Sixth, these do different things when the lists have different lengths. Using (==) directly on lists will result in False , whereas zipWith will ignore the excess elements. That is:

[1,2,3] == [1,2]                   -- False
and $ zipWith (==) [1,2,3] [1,2]   -- True

Now, there are probably situations when you want the second behaviour. But you almost certainly want the first behaviour.

Finally, to emphasise, just use (==) directly on the lists:

listA == listB

Answer 3

您可以替换all (\\x -> x)与and 。