Php array: Spot the difference!

Question

I can't figure out what the difference is between the final result of $arrayParams in this snippet of PHP.

The function takes string $types, and another parameter called $params. $params itself can be a single value of any type, or an array of values of any type.

The purpose of this code is for binding to call_user_func_array() later on in the code.

The final $arrayParams variable needs to be an array with index 0 as the original $types string, and then the following indices are to be references of the string(s) passed in as $params.

if(is_array($params)) {
  // Make a new array, first index is $types string.
  $arrayParams = array($types);

  // Loop over $params array and add the pointer of each index to $arrayParams.
  // ??? This doesn't seem to be working ???
  foreach($params as $p) {
    $arrayParams[] = &$p;
  }
 }
else {
  // This works fine here, very simple.
  $arrayParams = array($types, &$params);
}

The var_dump of $array_params, when passing $params as an array shows all keys after the first to be a pointer to the same value (?)

Stefan Gehrig · Accepted Answer · 2010-12-11 13:10:33Z

2

// ...
foreach($params as &$p) {
    $arrayParams[] = &$p;
}
unset($p);
// ...

foreach operates on a copy of the $params-array, unless you specify to use references on the iteration.

answered Dec 11, 2010 at 13:10

Stefan Gehrig

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3 Comments

Greg Over a year ago

Can I ask why $p needs to be unset?

Alin P. Over a year ago

@Greg See the warning here.

Greg Over a year ago

Accepted answer because of unset tip. Thanks :)

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