Edit report at https://bugs.php.net/bug.php?id=74302&edit=1
ID: 74302
Updated by: [email protected]
Reported by: xtpd17 at gmail dot com
Summary: yield from1251('dfd') ("from1251" is a valid
function name) fails
Status: Closed
Type: Bug
Package: *General Issues
Operating System: Windows 7
PHP Version: 7.1.3
Assigned To: pollita
Block user comment: N
Private report: N
New Comment:
btw, just for the record, you can work around this bug for now with the following:
yield (from1251('df'));
The extra parenthesis doesn't change the meaning, but it does force the lexer to not consume the "from" portion of the function name.
Previous Comments:
------------------------------------------------------------------------
[2017-03-23 20:51:10] [email protected]
https://github.com/php/php-src/commit/0fb640c71763ddb1b8017c87cec10fc76764feff
------------------------------------------------------------------------
[2017-03-23 19:40:48] [email protected]
Verified. Fix should be simple enough.
------------------------------------------------------------------------
[2017-03-23 19:10:54] xtpd17 at gmail dot com
Description:
------------
Seems like parser thinks that is a "yield from generator" case, which is not.
Test script:
---------------
<?php
function from1251($a)
{
return $a;
}
function foo()
{
yield from1251('df');
}
// Parse error: syntax error, unexpected '(' in test.php on line 10
------------------------------------------------------------------------
--
Edit this bug report at https://bugs.php.net/bug.php?id=74302&edit=1
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