Hi!
Why function can't return references to nothing?
In some cases wound be useful return nothing.
Here I want show example of code which implements search leaf in tree
and if no found path then return nothing and this correctly works.
#!/usr/bin/php
<?php
function &getmultivalue(&$arr, $keys) {
//Begin from root
$value = &$arr;
for ($i = 0; $i < count($keys); $i++) {
//if not found
if(!isset($value[$keys[$i]])) return;
//Next leaf
$value = &$value[$keys[$i]];
}
return $value;
}
$tree = ['aaa'=>['bbb'=>['ccc'=>'ddd']]];
print_r($tree);
$found = &getmultivalue($tree, ['aaa', 'bbb', 'ccc']);
if(isset($found)) {
$found = 'zzz';
}
$found = &getmultivalue($tree, ['aaa', 'bbb', 'ddd']);
if(isset($found)) {
$found = '---';
}
print_r($tree);
?>
=======================
$ ./example.php
Array
(
[aaa] => Array
(
[bbb] => Array
(
[ccc] => ddd
)
)
)
PHP Notice: Only variable references should be returned by reference
in /home/mikhail/example.php on line 8
Array
(
[aaa] => Array
(
[bbb] => Array
(
[ccc] => zzz
)
)
)
Work as expected but I really do not like Notice "Only variable
references should be returned by reference" which said that is my code
is ugly.
Can you suggest alternative solution or may be remove this notice in this case.
Of course I am agree that impossible return constants, but i think
nothing we could return.
function &getValue() {
return 'value'; // this is not right.
}
but
function &getValue() {
return; // this is right.
}
What do you think?
Sorry if I did write to wrong mailing list. I would like that it read
peoples who develop PHP standards.
--
Best Regards,
Mike Gavrilov.
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