ID: 10967
Updated by: jeroen
Reported By: [EMAIL PROTECTED]
Old Status: Open
Status: Bogus
Bug Type: Scripting Engine problem
Operating System: win2k
PHP Version: 4.0.5
New Comment:
This is nonsense, what do you expect of $x .=
&somefunction()? that the second part of $x gets referenced?
Previous Comments:
------------------------------------------------------------------------
[2001-06-12 15:36:34] [EMAIL PROTECTED]
well your playing with references where they are not needed.. expect to get your
fingers burnt.
------------------------------------------------------------------------
[2001-05-22 16:50:28] [EMAIL PROTECTED]
uhm, well, the thing with the $temp var is useless. i see now that i cannot reference
something into *a part* of something else.
but the silent loss of "<br>\n" is still a problem, imo.
fab
------------------------------------------------------------------------
[2001-05-18 23:41:12] [EMAIL PROTECTED]
code i would like to use:
---cut---
function &someShit() {
return 'foo';
}
$out = '';
for ($i = 1; $i <= 3; $i++) {
$out .= &someShit() . "<br>\n";
}
echo $out;
---cut---
problem:
parse error on line
$out .= &someShit() . "<br>\n";
because .= and & don't work together.
so the workaround would be:
$temp = &someShit() . "<br>\n";
$out .= $temp;
problem here:
it prints out 'foofoofoo' and not
'foo<br>\nfoo<br>\nfoo<br>\n'
so the code finally looks like:
---cut---
function &someShit() {
return 'foo';
}
$out = '';
for ($i = 1; $i <= 3; $i++) {
$temp = &someShit();
$out .= $temp . "<br>\n";
}
echo $out;
---cut---
is this the normal behavior?
fab
------------------------------------------------------------------------
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