对算术运算符和函数之间的行为感到困惑？

Question

function pow(x, n) {
  if (n == 1) {
    return x;
  } else {
    return x * pow(x, n - 1);
  }
}

alert( pow(2, 3) ); // 8

source = https://javascript.info/recursion

Hello all: I'm confused about the second return statement of this function:

return x * pow(x, n - 1);

I'm just looking for either some clarification or a reference to that behavior.

From my perspective, it looks like x is multiplied only by the first parameter of the function, and the n-1 is ignored.

(How does n-1 affect the result <- original question)

Sorry, I messed up the original question... I want to ask how does javascript interprets that multiplication. When multiplying an integer and a function, I don't quite understand what's happening. How does javascript choose what to multiply with more than one parameter?

Answer 1

pow(2, 3) = 2 * pow(2, 2) = 2 * 2 * pow(2, 1) = 2 * 2 * 2

You are not actually calculating a product with n - 1, but refer to n as a counter. This is equivalent to

var result = 1;
while (n >= 0) {
    result *= x;
    n--;
}