error with &$variable

[artev]

Parse error: syntax error, unexpected '=', expecting ')'

on function so

function pippo($var1, &$var2 = null, $var3 = null) { ....}

but if remove &
I not have the erros;

I must change sometime in php.ini or can resolve how?
I use php4.x

[=?iso-8859-1?Q?=C1lvaro?= G. Vicario]

*** artev escribió/wrote (Tue, 12 Aug 2008 23:30:42 +0200):

function pippo($var1, &$var2 = null, $var3 = null) { ....}

but if remove &
I not have the erros;

I must change sometime in php.ini or can resolve how?
I use php4.x

According to the manual, you must upgrade to PHP 5 or rewrite your
function. PHP 4 doesn't allow the combination of passing by reference and
default values:

http://es2.php.net/manual/en/functions.arguments.php
--
-- http://alvaro.es - Álvaro G. Vicario - Burgos, Spain
-- Mi sitio sobre programación web: http://bits.demogracia.com
-- Mi web de humor en cubitos: http://www.demogracia.com
--

[artev]

According to the manual, you must upgrade to PHP 5 or rewrite your
function. PHP 4 doesn't allow the combination of passing by reference and
default values:

http://es2.php.net/manual/en/functions.arguments.php

ok so this work only in php5
function pippo($var1, &$var2 = null, $var3 = null) { ....}

but if I write so:
function pippo($var1, &$var2, $var3 = null)
{
if(!&$var2) {&$var2 = null}
.....
}

I can resolve?

[Jerry Stuckle]

artev wrote:

>According to the manual, you must upgrade to PHP 5 or rewrite your
function. PHP 4 doesn't allow the combination of passing by reference and
default values:

http://es2.php.net/manual/en/functions.arguments.php

ok so this work only in php5
function pippo($var1, &$var2 = null, $var3 = null) { ....}

but if I write so:
function pippo($var1, &$var2, $var3 = null)
{
if(!&$var2) {&$var2 = null}
....
}

I can resolve?

if (!$var2) {$var2 = null;}

--
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
js*******@attglobal.net
==================

[=?iso-8859-1?Q?=C1lvaro?= G. Vicario]

*** artev escribió/wrote (Wed, 13 Aug 2008 17:18:43 +0200):

ok so this work only in php5
function pippo($var1, &$var2 = null, $var3 = null) { ....}

but if I write so:
function pippo($var1, &$var2, $var3 = null)
{
if(!&$var2) {&$var2 = null}
....
}

It's not exactly the same. This way you must call pippo() with at least
$var1 and $var2, so $var2 is no longer optional.

I can think of a couple of tricks but it looks like that the best solution
would be a small redesign.

--
-- http://alvaro.es - Álvaro G. Vicario - Burgos, Spain
-- Mi sitio sobre programación web: http://bits.demogracia.com
-- Mi web de humor en cubitos: http://www.demogracia.com
--

[artev]

Il Wed, 13 Aug 2008 18:14:32 +0200, Álvaro G. Vicario ha scritto:

*** artev escribió/wrote (Wed, 13 Aug 2008 17:18:43 +0200):

>ok so this work only in php5
function pippo($var1, &$var2 = null, $var3 = null) { ....}

but if I write so:
function pippo($var1, &$var2, $var3 = null)
{
if(!&$var2) {&$var2 = null}
....
}

It's not exactly the same. This way you must call pippo() with at least
$var1 and $var2, so $var2 is no longer optional.

function pippo($var1, &$var2 = null, $var3 = null) { ....}
but also so $var2 is no longer optional ?

[Michael Fesser]

..oO(artev)

>According to the manual, you must upgrade to PHP 5 or rewrite your
function. PHP 4 doesn't allow the combination of passing by reference and
default values:

http://es2.php.net/manual/en/functions.arguments.php

ok so this work only in php5
function pippo($var1, &$var2 = null, $var3 = null) { ....}

but if I write so:
function pippo($var1, &$var2, $var3 = null)
{
if(!&$var2) {&$var2 = null}
....
}

I can resolve?

Upgrade to PHP 5.

Micha