Would you consider this a bug, or at least unexpected behaviour?
php 4.3.10 windows, and 4.3.9 nixxy both give the same output
function &gooGoo()
{
static $y = 'Y Static Initial val';
echo "gooGoo Y: $y<br>";
return $y;
}
$foo = &gooGoo(); $foo = 'Change one'; // yes, expect change
$foo = gooGoo(); $foo = 'Second change'; // don't expect change - but there IS
$baz = gooGoo(); $baz = 'Third go'; // don't expect change - and there isn't
gooGoo();
Its expected since you are returning by reference...
It may be a little clearer like this, i think
<?php
function &gooGoo()
{
static $y = 'Y Static Initial val';
echo "gooGoo Y: $y<br>";
return $y;
}
$foo = &gooGoo(); //return reference of $y to $foo
$foo = 'Change one'; //foo points to the same address as $y,
//so the value "changes" for both
$foo = gooGoo(); //reference still exists, technically you dont even
//need the assignment. gooGoo() would do the same
$foo = 'Second change'; //foo still points to address of $y
$baz = gooGoo(); //$y = 'Second change' return this value to $baz
$baz = 'Third go'; //$baz does not point to $y or $foo so they wont change
//if you did $baz =& gooGoo(); it would change the reference
gooGoo(); //no changes in reference. $y and $foo still point to same place
?>
Here's my interpretation:
Basically the ampersand must be used in both the function definition and function call for return by reference to work as expected.
Originally posted by laserlight
Here's my interpretation:
- In the 2nd call to gooGoo(), 'Change one' is printed since that's the value of $y.
$foo = gooGoo() effectively assigns $y to $foo - but $foo already is $y, so there's no net effect.Basically the ampersand must be used in both the function definition and function call for return by reference to work as expected. [/B]
My interpretation when I found this was the same as point 3. Really foo should be set to a COPY of $y, not a reference on its 2nd call as the ampersand isn't used, so the internal $y shouldn't change on $foo = 'second change', and this is illustrated when $baz is assigned.
I'm going to write another experiment that allows the changing of the static y with via a different path and see if it splits the reference count and what I would consider the same.
after you return by reference one time from a function to a variable, the reference remains in place.
i.e.
function &x()
{
static $x = 0;
++$x;
echo "func \$x = $x<br />\n";
return $x;
}
$xx =& x(); //$xx = 1, $x = 1 $xx points to the same addr as $x
x(); //$xx = 2 $x = 2
x(); //$xx = 3 $x = 3
++$xx; //$xx = 4; $x = 4
x();
echo $xx; //outputs 5
Hi Drew. I know about the reference staying in place, but what was fooling me in my example was the second assignment to $foo without the ampersand. It looks like it shouldn't work, you're not assigning by reference the 2nd time, so how does a change of $foo change $y - this now explains it and puts the beast to sleep in my head, it was nothing to do with reference counting or anything that I tried explaining it with, essentially a 'trick of the light'.
<?php
function &gooGoo($returnString = 0)
{
static $yStatic = 'Y Static Initial val';
echo "gooGoo Y: $yStatic <br>";
if ($returnString)
{
echo ' you want string, you get string<br />';
$yStatic = 'change by string condition';
return 'Hi, I am returned string';
}
else
return $yStatic;
}
$proxy = &gooGoo(); // get a control reference to $yStatic
$foo = &gooGoo(); // $foo a definite reference to $yStatic
$foo = 'Change one'; // so $yStatic = 'Change one';
$foo = gooGoo(1);
/* Above I now know $foo equals 'Hi, I am a returned string'
and I assume $yStatic now to be 'change by string condition'
but check it out and it's NOT.
$foo = gooGoo(1) the 2nd time essentially behaves as
$yStatic = 'Hi, I am a returned string' because
$foo IS $yStatic - the eye is just fooled by $foo = gooGoo(1);
*/
echo "\$foo = $foo<br />";
echo '\$proxy = ', $proxy, '<br />';
$foo = 'Second change'; // surely no change - but there IS [though now understood]
$baz = gooGoo();
$baz = 'Third go'; // don't expect change - and there isn't, and never was
gooGoo();
?>