note 64983 deleted from language.types.type-juggling by bjori

Note Submitter: Reason: bad code ---- There seems to be a problem with string hexadecimals and type juggling. The following expectedly ignores the hexadecimal value since floats are only specifiable in decimal. <?php var_dump ((float) "0x31"); // (float)0 var_dump (1.5+(float)"0x31"); // (float)1.5 ?> However, the following seems to first convert the hexadecimal string to an integer (which does allow hexadecimals) and then to a float, rather than immediately converting to float, as the documentation here on type juggling would have it: <?php var_dump (1.5+"0x31"); // (int)50.5 (i.e., 1.5+49) ?>