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Recursive function problem

Posted on 2006-11-19
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Last Modified: 2006-11-19
function test($a) {
 if ($a > 10) {
  if ($a > 50) {
   test($a-10);
  }
  else {
   test($a-5);
  }
 }
 else {
  return $a;
 }
}

var_dump(test(100));

The above "example" function returns NULL.

Where am i going wrong?

Thank you in advance.
0
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Question by:fox_km
  • 2
  • 2
4 Comments
 
LVL 49

Accepted Solution

by:
Roonaan earned 125 total points
ID: 17974180
Use:

function test($a) {
 if ($a > 10) {
  if ($a > 50) {
   return test($a-10);
  }
  else {
   return test($a-5);
  }
 }
 else {
  return $a;
 }
}

-r-
0
 

Author Comment

by:fox_km
ID: 17974221
Thank you.

Could you possibly explain why this works?

My logic was, you only want to return something if $a is less then 10. I'm assuming the extra returns serve the purpose of halting the execution of the current function call. I had assumed that it wouldn't matter, as those calls do not return anything, therefore the original function call would only have a value returned when $a <= 10. Is my understanding of recursive functions incorrect?
0
 
LVL 49

Expert Comment

by:Roonaan
ID: 17974232
Well you would either return it, or store the subfunctions call into a:

function test($a) {
 if ($a > 10) {
  if ($a > 50) {
   $a = test($a-10);
  }
  else {
   $a = test($a-5);
  }
 }
 return $a;
}

When you just call text($a - 10) and do nothing with its result, there is no sense in calling the function at all.

-r-
0
 

Author Comment

by:fox_km
ID: 17974242
Okay, i understand. Thanks again.
0

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