recursion and function call in javascript

Question

function test(param, n) {
  console.log(param, "start --- ", n);

  if (n > 1) {
    console.log(param, "before first recursion --- ", n);

    test("test-1", n - 1);
    console.log(param, "after first recursion --- ", n);

    test("test-2", n - 1);
    console.log(param, "after second recursion --- ", n);
  }

  console.log(param, "end --- ", n);
}

test("initialize", 3);

In output console

test-1 after first recursion --- I am getting value 2, can someone please explain the flow of code is going as, I was expecting 3 as the output, because in my understanding if the test("test-1", n-1) is completed then the next call i.e. test("test-2", n-1) will be getting 3 value as fresh input, also is the js working synchronous here ?

It's fully synchronous. It's just executing the recursion before continuing. — VLAZ
– VLAZ, Commented Jan 2, 2020 at 8:40
Your output should be: initialize start --- 3, initialize, before first recursion --- 3, test-1 start --- 2, test-1 before first recursion --- 2, test-1 after first recursion --- 2, test-2 start --- 1, test-2, end --- 1 — SPlatten
– SPlatten, Commented Jan 2, 2020 at 8:44

VLAZ · Accepted Answer · 2020-01-02 09:11:32Z

2

You could use some indentations to the output and look which level is actually running.

function test(param, n, level = 0) {
    console.log(''.padStart(level * 2), level, param, "start --- ", n);
    if (n > 1) {
        test("test-1", n - 1, level + 1);
        test("test-2", n - 1, level + 1);
    }
    console.log(''.padStart(level * 2), level, param, "end --- ", n);
}

test("initialize", 3);

edited Jan 2, 2020 at 9:11

VLAZ

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answered Jan 2, 2020 at 8:43

Nina Scholz

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