Haskell foldl函数疑问：累加器参数传递机制解析

Great question! Let's break down exactly how foldl routes its arguments to your lambda function, using your sum' example to make it concrete.

First, let's recap the definition of foldl itself—its type signature tells us a lot:

foldl :: (b -> a -> b) -> b -> [a] -> b

Breaking this down:

• The first argument is a folding function (your lambda \acc x -> acc + x) that takes an accumulator (b) and a list element (a), then returns a new accumulator (b).
• The second argument is the initial accumulator value (0 in your case).
• The third argument is the list we're folding over ([3,5,2,1]).

Step-by-Step Execution of Your sum' Example #

When you run sum' [3,5,2,1], here's how foldl handles the parameter passing:

1. First iteration: foldl takes the initial accumulator value 0 and passes it as the acc parameter to your lambda. It then takes the first element of the list, 3, and passes it as the x parameter. The lambda computes 0 + 3 = 3—this becomes the new accumulator for the next step.
2. Second iteration: Now the accumulator is 3, and the next list element is 5. The lambda runs 3 + 5 = 8, updating the accumulator again.
3. Third iteration: Accumulator is 8, list element is 2 → 8 + 2 = 10.
4. Fourth iteration: Accumulator is 10, list element is 1 → 10 + 1 = 11.
5. Since there are no more elements in the list, foldl returns the final accumulator value 11.

A Simplified Way to Visualize It #

You can think of foldl expanding into a nested sequence of function calls. For your example:

foldl (\acc x -> acc + x) 0 [3,5,2,1]
-- Is equivalent to:
(\acc x -> acc + x) 
  ((\acc x -> acc + x) 
    ((\acc x -> acc + x) 
      ((\acc x -> acc + x) 0 3) 
    5) 
  2) 
1

Each time, the result of the inner call becomes the acc for the outer call, and the next list element becomes x. That's how 0 and 3 end up as the first set of parameters—foldl starts by pairing the initial value with the first list element, then chains the results.

To put it plainly: foldl is designed to feed the initial value to your folding function first, then work through the list elements one by one, using each result as the accumulator for the next step.

内容的提问来源于stack exchange，提问作者ceno980