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PHP Functional programming

Discussion in 'PHP' started by keenlearner, Dec 31, 2007.

0
  1. #1
    I need to do a few functional programming in PHP, here is the code but it only prints out "arg2" but not "arg1". Thank you.

    
function prints($arg1, $arg2)
{
  echo $arg1. $arg2;
}


function applyOne($arg1)
{
  $f =  create_function('$arg2', 'return prints("$arg1", $arg2);');
  return $f;
}

$printFunction = applyOne("arg1");

$printFunction('arg2');
    PHP:

     
    keenlearner, Dec 31, 2007 IP
  2. nico_swd

    nico_swd Prominent Member

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    #2
    $arg2 is unset in your applyOne function. What do you want it to output?
     
    nico_swd, Dec 31, 2007 IP
  3. keenlearner

    keenlearner Peon

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    #3
    I want it to output 'arg1arg2';
     
    keenlearner, Dec 31, 2007 IP