Bug #74302 [Com]: yield from1251('dfd') ("from1251" is a valid function name) fails

Edit report at https://bugs.php.net/bug.php?id=74302&edit=1 ID: 74302 Comment by: xtpd17 at gmail dot com 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 Yep, thanks, that was the way I fixed it at first almost automatically and calmed down ("well, it's just my migration from 5.6 and new stricter parenthesis rules in 7.1"), but it looked strange enough to dig into it a bit just out of curiosity )). Previous Comments: ------------------------------------------------------------------------ [2017-03-23 23:49:06] pollita@php.net 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. ------------------------------------------------------------------------ [2017-03-23 20:51:10] pollita@php.net https://github.com/php/php-src/commit/0fb640c71763ddb1b8017c87cec10fc76764feff ------------------------------------------------------------------------ [2017-03-23 19:40:48] pollita@php.net 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