How does empty return statement in conditional statement functions?

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Learning JS from MDN Docs came across code in “functions section”. Not able to understand what does return; in the below code accomplish

function foo(i) {
  if (i < 0)
    return; // what does this do? 
  console.log('begin: ' + i);
  foo(i - 1);
  console.log('end: ' + i);
}
foo(3);

Output

'begin: 3'
'begin: 2'
'begin: 1'
'begin: 0'
'end: 0'
'end: 1' // why are these printing
'end: 2' // this one
'end: 3' // this one

I understood first 5 lines of output, but not able to understand why end: 0,1,2,3 are coming?

Please help !

Answer

return terminates the current function, and returns control flow to the caller.

When foo(3); is called, the function is initialized with an i parameter of 3. It fails the if statement, prints begin: 3, and then calls foo(3 - 1);. At this point, the current function (the one with an i parameter of 3) is paused until the foo(3 - 1); call completes.

foo(2); prints begin: 2, and then pauses while calling foo(1).

foo(1) prints begin: 1, and then pauses while calling foo(0).

foo(0) prints begin: 0, and returns: it terminates, and yields control flow back to its caller, the function call of foo(1).

foo(1) resumes, and continues executing, printing end: 1. That’s the end of the function block, so the foo(1) function call ends, yielding control flow back to the foo(2) function call.

foo(2) resumes, printing end: 2, then terminates, yielding control flow back to foo(3). Then foo(3) prints end: 3, and terminates.


return only terminates the current function. The only way to terminate all calling functions (until a catch is encountered) would be to throw an error:

function foo(i) {
  if (i < 0)
    throw new Error();
  console.log('begin: ' + i);
  foo(i - 1);
  console.log('end: ' + i);
}
foo(3);


Source: stackoverflow