咖喱形式的递归函数可以是尾递归吗？

Question

Given is the following recursive map function:

 const U = f => f(f); const map = f => U(h => acc => ([head, ...tail]) => head === undefined ? acc : h(h)([...acc, f(head)])(tail))([]); const xs = [1,2,3,4,5]; console.log(map(x => x * x)([1,2,3,4,5])); 

Obviously, the recursive call h(h) isn't the last action of the recursive function. But when the stack is unwound, everything that happens is that the finished accumulator is returned without any further changes or operations. Is map against my expectations tail recursive?

Answer 1

the recursive call h(h) isn't the last action of the recursive function

h(h) is not a recursive call. The …(tail) is the recursive call, and yes, it's in a tail position.

This might get more obvious if you drop that overcomplicated U combinator (or at least used the Y combinator right away):

const map = f => {
  const mapF = acc => ([head, ...tail]) => head === undefined
    ? acc
    : mapF([...acc, f(head)])(tail);
  return mapF([]);
};