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Improving error messages when using indirect references

php.internals

Martin Jansen

22 years ago

Hey,

<?php
error_reporting(E_ALL);

echo $$foo;
?>

results in

Notice: Undefined variable: foo in /foo/bar.php on line 4

Notice: Undefined variable: in /foo/bar.php on line 4

Wouldn't it be more appropriate if the second notice was something like
"Notice: Undefined variable: <unnamed> in /foo/bar.php on line 4"?

--
- Martin                           Martin Jansen
http://martinjansen.com/

Hartmut Holzgraefe

22 years ago

Martin Jansen wrote:

> Hey,
>
>         <?php
>         error_reporting(E_ALL);
>
>         echo $$foo;
>         ?>
>
> results in
>
>         Notice: Undefined variable: foo in /foo/bar.php on line 4
>
>         Notice: Undefined variable: in /foo/bar.php on line 4
>
> Wouldn't it be more appropriate if the second notice was something like
> "Notice: Undefined variable: <unnamed> in /foo/bar.php on line 4"?
>

i'd prefere to put the variable name in single quotes here
as the name actually is '' (empty string)

<?php
$foo = "";
$$foo = "hallo";
echo ${""};
?>

--
Hartmut Holzgraefe  <hartmut@php.net>

Derick Rethans

22 years ago

On Fri, 13 Feb 2004, Martin Jansen wrote:

> Hey,
>
>         <?php
>         error_reporting(E_ALL);
>
>         echo $$foo;
>         ?>
>
> results in
>
>         Notice: Undefined variable: foo in /foo/bar.php on line 4
>
>         Notice: Undefined variable: in /foo/bar.php on line 4
>
> Wouldn't it be more appropriate if the second notice was something like
> "Notice: Undefined variable: <unnamed> in /foo/bar.php on line 4"?

Would be a lie... $foo WILL evaluate to the empty string which used like
this, although it does look weird, it is right :)

Derick