php - How to set optional parameter to default without passing it?

Question

How can i skip first argument without giving any value in function call,so that first argument can take default value NULL?

function test($a = NULL, $b = true){
  if( $a == NULL){
     if($b == true){
       echo 'a = NULL, b = true';
     }
     else{
       echo ' a = NULL, b = false';
     }
  }
  else{
    if($b == true){
      echo 'a != NULL, b = true';
    }
  else{
      echo ' a!=NULL, b = false';
      }
  }
}

test();        //ok
test(NULL);    //ok
test(5);       //ok
test(5,false)  //ok
test(,false);  // How to skip first argument without passing any value? ->PARSE error

// i don' want to use default value for first argument, although test(NULL,false)
// or test('',false) will work but can i skip first argument somehow? 
// i want to pass only second argument, so that first arg will be default set by
// function

Answer 1

You cannot just skip an argument in PHP.

You may wish to consider the Perl trick and use an associated array. Use array_merge to merge the parameters with the defaults.

e.g.

function Test($parameters = null)
{
   $defaults = array('color' => 'red', 'otherparm' => 5);
   if ($parameters == null)
   {
      $parameters = $defaults;
   }
   else
   {
      $parameters = array_merge($defaults, $parameters);
   }
 }

Then call the function like this

Test(array('otherparm' => 7));