Understanding how a recursive function works in javascript

Question

I am a little bit confused on the exact execution timeline in which the following function runs. I saw this code on the MDN guide:

function foo (i)  {
    if (i<0) return;
    console.log('begin' + i);
    foo(i-1);
    console.log('end' + i);
}

Based on my knowledge of how functions work, this is what I thought, though my knowledge is inadequate:

I think when the function foo(3) is called, it will go to the function and check the condition if i < 0, then it will run the next code console.log('begin:' + i).

After that, the next line of code is executed, since JavaScript execute line by line, the next line of code which is foo(i-1) will be executed.

It takes the current value of i which is 3 minus 1, so foo(2) is called and the execution continues.

My question is: Is my understanding correct?. If no please explain what this code is doing, otherwise can anyone explain breiefly when will the function stop being called?

foo(3) begin 3 begin 2 begin 1 begin 0 end 0 end 1 end 2 end 3 — BNilsou
– BNilsou, Commented Jul 12, 2016 at 7:30
From your description you do understand how this recursive function works! — Jamiec
– Jamiec, Commented Jul 12, 2016 at 7:31
I was editing your question to improve readability, then I found actually you are not asking anything ... — shole
– shole, Commented Jul 12, 2016 at 7:35

Angelos Chalaris · Accepted Answer · 2016-07-12 07:35:10Z

2

What happens when you call foo(3) is as follows (in pseudocode):

execute foo(3)
    3 is greater than 0 thus log(begin 3)
    execute foo(2)
        2 is greater than 0 thus log(begin 2)
        execute foo(1)
            1 is greater than 0 thus log(begin 1)
            execute foo(0)
                0 is equal to 0 thus log(begin 0)
                execute foo(-1)
                    -1 is less than 0 thus return
                log(end 0)
            log(end 1)
        log(end 2)
    log(end 3)
return

answered Jul 12, 2016 at 7:35

Angelos Chalaris

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2 Comments

pedroyanky Over a year ago

pls log(end 0) log(end 1) log(end 2) log(end 3) return what conditions led to the printing out of those .

Angelos Chalaris Over a year ago

@pedroyanky no conditions led to those. When the function that is called before them (eg foo(0)) returns, the calling function will resume execution so it will run the next line of code, which is a log(end ...). Think of the foo(number) call inside the foo() function as a nother line of code that could be anything else (like for example one line like console.log(' '). Hope this helps you understand how recursion works!

Xat_MassacrE · Accepted Answer · 2016-07-12 07:38:40Z

1

It works like this:

foo(3)->

if (3 < 0)
  return
console.log('begin' + 3)
if (2 < 0)
  return
console.log('begin' + 2)
if (1 < 0)
  return
console.log('begin' + 1)
if (0 < 0)
  return
console.log('begin' + 0)
if (-1 < 0)
  return
console.log('end' + 0)
console.log('end' + 1)
console.log('end' + 2)
console.log('end' + 3)

answered Jul 12, 2016 at 7:38

Xat_MassacrE

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