Bug #74302 [Csd]: 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 Updated by: pollita@php.net 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] 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