Array issue within function

Question

I have:

<?php
$a=array('x'=>3,'y'=>6,'z'=>12);  //NOTE THIS*** position 1
echo func(5);

function func($c)
{
  $a = array('x'=>3,'y'=>6,'z'=>12);  //NOTE THIS*** position 2
  $previous = null;
  foreach($a as $k => $v)
  {
     if($v > $c) // This part was unclear, so it could be >= instead
     {
        return $previous;
     }
     $previous = $k;
  }
  return $previous;
}

Now, when I have the array $a inside the function (position 2), it works perfect. However, when I place $a outside the function (position 1) it does not work.

Why is this?

Jacob · Accepted Answer · 2011-03-07 22:36:00Z

2

Functions can only access data within their scope.

$a defined in position one is in the global scope, if you want to access that in the function you need to pass it as an argument to the function.

Or you can add the line in the function

global $a;

Which brings $a into the current scope from the global scope.

Have a read of Variable Scope from the PHP documentation.

answered Mar 7, 2011 at 22:36

Jacob

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4 Comments

David19801 Over a year ago

This is something I had not come accross in other programming languages. Thanks for the detail!

user438293456 Over a year ago

@David19801 Variable Scope is an important concept in programming. If you don't know that, I would also point to keithjgrant's answer where he mentions that using global variables is a generally frowned upon.

Jacob Over a year ago

@swiftee they are treated differently in PHP. Using a global variable in C, you can access it in the function without doing anything explicitly to bring it into the function. Same with Javascript. Although obviously scope still exists. I'm not advocating global variables in any sense tho, hence including the option of passing it as an argument in my answer.

user438293456 Over a year ago

I wasn't trying to imply that you were advocating it. :) I just wanted to nudge the OP towards the first option.

boo-urns · Accepted Answer · 2011-03-07 22:36:35Z

2

This has to do with variable scope in PHP. You can check that here: http://php.net/manual/en/language.variables.scope.php

Declaring $a in position 1 makes it global. In the way you want to access it, you need to use the global keyword: global $a;

answered Mar 7, 2011 at 22:36

boo-urns

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1 Comment

user438293456 Over a year ago

+1 for first to attach documentation; other answers followed suit.

keithjgrant · Accepted Answer · 2011-03-07 22:36:34Z

1

The $a at position one is "out of scope". A function does not have access to variables declared outside of it, unless specifically passed in as a parameter.

See PHP variable scope.

You can give the function access via the global keyword:

$a = array();

function myFunc() {
   global $a;
   // do something with $a here
}

... but note that the use of global variables is generally considered bad practice and is frowned upon.

answered Mar 7, 2011 at 22:36

keithjgrant

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Comments

Jumpey · Accepted Answer · 2011-03-07 22:35:44Z

0

PHP can not use global variables inside functions unless you declare them:

function func($c)
{
  global $a;
  # The rest of the code.
}

answered Mar 7, 2011 at 22:35

Jumpey

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