Return doesn't exit recursive function in javascript

Question

I've read the dozen variations on this question, but those answers haven't led me to what must be an obvious mistake. Why does this always return false? Why do I see called again even after a found it? And if I put a return in front of the recursive call, why do I never see found it?

function subResult (object, start, target){
    console.log('called again')
    if (start === target){
      console.log('found it')
      return true
    } else {
      for (var i = 0; i < object[start].edges.length; i++){
        subResult(object, object[start].edges[i], target)
      }
    }
   return false
 }

Ed Heal · Accepted Answer · 2016-04-23 07:09:01Z

4

Change

for (var i = 0; i < object[start].edges.length; i++){
    subResult(object, object[start].edges[i], target)
}

to

for (var i = 0; i < object[start].edges.length; i++){
    if (subResult(object, object[start].edges[i], target)) {
       return true;
    }
}

I.e. when found your done. If not keep going.

answered Apr 23, 2016 at 7:09

Ed Heal

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1 Comment

rswerve Over a year ago

Wow, that was it. Thanks so much. So I'm saying something like "if the recursive call returns true, return true"? I really thought it would be sufficient to do the first return true and then return the recursive call (as in Abu Hanifa's answer) but that just resulted in false always being returned.

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