In PHP, I have
error_reporting set to report everything including notices.
Why does the following not throw any notices, errors or anything else?
$myarray = null; $myvalue = $myarray['banana'];
$myarray = array(); $myvalue = $myarray['banana']; // throws a notice, as expected ✔ $myarray = (array)null; $myvalue = $myarray['banana']; // throws a notice, as expected ✔ $myarray = null; $myvalue = $myarray['banana']; // no notice or warning thrown, $myvalue is now NULL. ✘ Why?
It’s possible it’s a bug in PHP, or I’m just not understanding something about how this works.
There is an active bug report started at 2006.
And in documentation it is a notice about this in String section.
There are three types which it might be valid to use the array derefence syntax on:
- Strings (to access the character at the given position)
- Object (objects implementing the ArrayAccess interface)
For all other types, PHP just returns the undefined variable.
As you can see, there is a special case for NULLs, and a default case, both returning the undefined variable.
Usually, when you try to use a value of one type as if it were another type, either an error or warning gets thrown or “type juggling” takes place. For example, if you try to concatenate two numbers with
., they’ll both get coerced to strings and concatenated.
However, as explained on the manual page about type juggling, this isn’t the case when treating a non-array like an array:
The behaviour of an automatic conversion to array is currently undefined.
In practice, the behaviour that happens when this “undefined behaviour” is triggered by dereferencing a non-array is that
null gets returned, as you’ve observed. This doesn’t just affect nulls – you’ll also get
null if you try to dereference a number or a resource.