Question

I have a function that takes a variable as a reference:

function get_articles($limit = 10, &$more = false){
  $results = get_results_from_db($limit); 
  $more = ($results->found > $limit) ? $results->found : false;
  return $results->data;
}

which I use it like:

$articles = get_articles(10, $more_results);

foreach($articles as $article){
  // do stuff
}

if($more_results) // we have more than 10 results

But I don’t get a notice telling me that $more_results above is undefined…
Is this normal?

Answer 1

Yes, because $more_results will always be defined at least with false value (after calling function).
Inside function variable will not be defined, but since you doesn’t read content of variable, notice is not generated.