为什么递归在两个例子中表现不同？

Question

I have two examples of recursion, one calculate the sum of all elements of the array and the second return the count, both examples behave differently and i can't relate!

First Example:

 const sum = (list) => { if (list.length === 0) { return 0; } return list[0] + sum(list.slice(1)); }; console.log(sum([1, 2, 3, 4])); // 10

The second:

 const count = (list) => { if (list.length === 0) { return 0; } return 1 + count(list.slice(1)); }; console.log(count([0, 1, 2, 3, 4, 5])); // 6

Why the first recursion go through all array elements adding every element while the other added 1 to each element then returned just the final value?? i assumed it would do the same and the difference would be just adding 1 to the sum!!

Answer 1

Look at return 1 + count(list.slice(1)); in the count function. It would just ignore the 1st element in the list you pass recursively to the function and always use 1. Versus the sum function does consider that element. That effectively returns the count of times the function has been called which is 6

Answer 2

1st - sum，2nd - count，如果你希望他们做同样的事情，你应该只使用其中的一个。

Answer 3

The recursion is the same.

It visits all elements of the array and returns zero if no element is available.

For each element, it returns either the item for summing or one for counting.