CODEWITHSUNDEEP
javascriptfunctionrecursion
Rabindra Kumar Mahato
Rabindra Kumar Mahato

Reputation: 99

js recursion function output question, function stack

I have confusion on function recursion

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

why is the output:

checking: 1
begin: 1
checking: 0
begin: 0
checking: -1
end: 0
end: 1

it should not print the end: 0, end: 1, because we are returning in if condition. What is happening?

Upvotes: 2

Views: 42

Answers (1)

CertainPerformance
CertainPerformance

Reputation: 370789

return only terminates the most immediately running function.

foo(1)
  foo(0)
    foo(-1)

The

if (i < 0)
  return;

gets fulfilled on the third call, terminating the third function

foo(1)
  foo(0)
    foo(-1) <-- this function stops running

but leaving the other functions to continue as normal:

foo(1)
  foo(0) (execution resumes here, just after the foo(1) line)

It's possible to get out of all ancestor functions (up until a .catch, if any) by throwing an error, but such a strategy just for the sake of control flow isn't a great idea.

function foo(i) {
  console.log("checking: " + i);
  if (i < 0)
    throw new Error();
  console.log('begin: ' + i);
  foo(i - 1);
  console.log('end: ' + i);
}
foo(1);

Upvotes: 4

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