Bug #10967 Updated: $x .= &someFunction();

From: Date: Tue, 12 Jun 2001 19:36:34 +0000
Subject: Bug #10967 Updated: $x .= &someFunction();
Groups: php.dev 
Request: Send a blank email to php-dev+get-57102@lists.php.net to get a copy of this message
ID: 10967 Updated by: jmoore Reported By: fabiankessler@usa.net Status: Open Bug Type: Scripting Engine problem Operating system: PHP Version: 4.0.5 Assigned To: Comments: well your playing with references where they are not needed.. expect to get your fingers burnt. Previous Comments: --------------------------------------------------------------------------- [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 --------------------------------------------------------------------------- [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 --------------------------------------------------------------------------- ATTENTION! Do NOT reply to this email! To reply, use the web interface found at http://bugs.php.net/?id=10967&edit=2

« previous php.dev (#57102) next »