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Posted by Jerry Stuckle on 11/18/68 11:45
Oli Filth wrote:
> Jerry Stuckle said the following on 09/04/2006 06:14:
>
>> Oli Filth wrote:
>>
>>> I have error level set to E_ALL | E_STRICT. The following code
>>> executes fine:
>>> <?php
>>>
>>> function getValueWithDefault(&$array, $key, $default = NULL)
>>> {
>>> return (isset($array[$key])) ? $array[$key] : $default;
>>> }
>>>
>>> $var = array();
>>> echo getValueWithDefault($var, "Roger", "Dodger") . "\n";
>>>
>>> ?>
>>>
>>> I see no reason why an error/warning should get thrown. isset()
>>> simply looks for an element with the specified key name in the
>>> associative array. In an empty array, the key doesn't exist, so
>>> isset() returns false.
>>>
>>
>> That's not what I said. I said if $var is EMPTY.
>
>
> From the PHP manual for empty():
>
> "The following things are considered to be empty:
> ...
> array() (an empty array)
> ..."
>
Yes, the array is empty. But $var is not! It contains an empty array.
Two different things.
>
>> Take out the $var=array() line and see what you get.
>
>
> So you mean "undefined"? Going back to the original purpose of this
> function/macro, why would you ever want to call it on an undefined array?
>
Because it may or may not exist, that's why.
> But anyway, removing the $var = array() line does not result in any
> error/warning. Because it's being passed by reference, PHP
> automatically creates a variable $var in the global scope (because for
> all it knows, this could be an output argument for the function).
>
>
Yep, and now you have changed the variable, haven't you?
--
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex@attglobal.net
==================
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