Help to see how it returns null

This post was last edited by Snowlove on 2013-06-17 18:27:34

 
  0 && $pos!=false) {$pos =newtripos ($str, $findstr, $count, $pos + 1);} Else{var_dump ($pos); return $pos;}} $a = "456123456455654466"; $b =newtripos ($a, ' 6 ', 4); Var_dump ($b);? >

Display $b after execution is null. And before the function returns, the Var_dump ($pos) is executed, and the display is int (16).
Why would var_dump ($b) be null?

Reply to discussion (solution)

You used recursion, return $pos; It will be used for $pos =newtripos ($str, $findstr, $count, $pos + 1) after entering recursion;
And you're not dealing with $pos in this branch (that is, returning the results to the upper-level recursion)
I may not be able to understand, see the code:

function Newtripos ($str, $findstr, $count, $off =0) {    $pos =stripos ($str, $findstr, $off);    $count--;    if ($count >0 && $pos!=false) {        $pos =newtripos ($str, $findstr, $count, $pos + 1);    }    return $pos;}

Your code is no problem.
function Newtripos ($str, $findstr, $count, $off =0) {
$pos =stripos ($str, $findstr, $off);
$count--;
if ($count >0 && $pos!=false) {
$pos =newtripos ($str, $findstr, $count, $pos + 1);
}else{
Var_dump ($pos);//The execution of the program here can correctly $pos the value. Continue execution should end function, return $pos
return $pos; This should return the value of the $pos.
}
}
$a = "456123456455654466";
$b =newtripos ($a, ' 6 ', 4); But here $b how not to receive $pos, is a null
Var_dump ($b);
?>

What's the difference between removing the else body and not removing it?

The difference is mostly because when else is not removed, the equivalent

 
  0 && $pos!=false) {$pos =newtripos ($str, $findstr, $count, $pos + 1);                return null;} Else{var_dump ($pos); return $pos;}}? >

And when you remove else, it is

 
  0 && $pos!=false) {$pos =newtripos ($str, $findstr, $count, $pos + 1);                return $pos;} Else{var_dump ($pos); return $pos;}}? >

