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Posted by Oli Filth on 06/18/35 11:45
Jerry Stuckle wrote:
> 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.
No, $var doesn't *contain* an empty array, $var *is* an empty array.
<?php
$var = array();
echo empty($var) ? "TRUE" : "FALSE"; // echoes "TRUE"
?>
> >> 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.
I thought the purpose was to check for the existence of an *entry* in
an associative array, such as $_POST, etc.
On the other hand, checking for the existence of a *variable* implies
bad practice. There should be no need. You should always know a
priori what variables exist at any given point in your code (assuming
the wonders of register_globals are disabled).
> > 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?
Only if it was undefined in the first place, which would be bad
practice.
--
Oli
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