How do foldl and foldr work, broken down in an example?

问题

Okay, I am new with scheme/racket/lisp. I am practicing creating my own functions, syntax, and recursion, so I want to make my own foldl and foldr functions that do exactly what the predefined versions do. I can't do it because I just don't understand how these functions work. I have seen similar questions on here but I still don't get it. Some examples broken down would help! Here is my (incorrect) process:

(foldl - 0 '(1 2 3 4)) I do 0 -(4-3-2-1) and get 2 which is the right answer

(foldl - 0 '(4 3 2 1)) I do 0-(1-2-3-4) and get 8 but it should be -2.

(foldr - 0 '(1 2 3 4)) I do 0-(1-2-3-4) and get 8 again, but it should be -2.

(foldr - 0 '(4 3 2 1)) I do 0-(4-3-2-1) and get 2 which is the right answer.

What am I doing wrong?

回答1:

Let's look at: (foldr - 0 '(1 2 3 4)).

Here the literal '(1 2 3 4) constructs a list whose elements are the numbers 1, 2, 3, and, 4. Let's make the construction of the list explicit:

(cons 1 (cons 2 (cons 3 (cons 4 empty))))

One can think of foldr as a function that replaces cons with a function f and empty with a value v.

Therefore

(foldr f 0 (cons 1 (cons 2 (cons 3 (cons 4 empty)))))

becomes

           (f 1    (f 2    (f 3    (f 4 v)))))

If the function f is - and the value v is 0, you will get:

(- 1 (- 2 (- 3 (- 4 0)))))

And we can calculate the result:

  (- 1 (- 2 (- 3 (- 4 0))))
= (- 1 (- 2 (- 3 4)))
= (- 1 (- 2 -1))
= (- 1 3)
= -2

Note that (foldr cons empty a-list) produces a copy of a-list.

The function foldl on the other hand uses the values from the other side:

> (foldl cons empty '(1 2 3 4))
'(4 3 2 1)

In other words:

(foldl f v '(1 2 3 4))

becomes

(f 4 (f 3 (f 2 (f 1 v)))).

If f is the function - and the value is 0, then we get:

  (- 4 (- 3 (- 2 (- 1 0))))
= (- 4 (- 3 (- 2 1)))
= (- 4 (- 3 1))
= (- 4 2)
= 2

Note that (foldl cons empty a-list) produces the reverse of a-list.

来源：https://stackoverflow.com/questions/42144068/how-do-foldl-and-foldr-work-broken-down-in-an-example