Bug #10967 Updated: $x .= &someFunction();
| From: | jeroen@php.net | Date: | Mon, 09 Jul 2001 12:37:50 +0000 |
| Subject: | Bug #10967 Updated: $x .= &someFunction(); | ||
| References: | 1 | Groups: | php.dev |
| Request: | Send a blank email to php-dev+get-59310@lists.php.net to get a copy of this message | ||
ID: 10967
Updated by: jeroen
Reported By: fabiankessler@usa.net
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:
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[2001-06-12 15:36:34] jmoore@php.net
well your playing with references where they are not needed.. expect to get your fingers burnt.
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[2001-05-22 16:50:28] fabiankessler@usa.net
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
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[2001-05-18 23:41:12] fabiankessler@usa.net
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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