RE: [PHP] Default value if parameter is not passed in

I notice that none of your variables use the PHP convention of $ preceding the variable name, I also do not see you defining a value for DEFAULT_VALUE, which by the upper case convention seems to be referring to a global constant. Is it not true (no pun intended) that if a variable (or constant) has not been defined, that assigning the contents of that variable (or value of the constant) will return a false (i.e. a 1)? Not sure I remember it all correctly but it seems to ring an ancient bell for me. HTH, Warren Vail -----Original Message----- From: Quanah Gibson-Mount [mailto:quanah@stanford.edu] Sent: Wednesday, October 27, 2004 4:35 PM To: php-general@lists.php.net Subject: [PHP] Default value if parameter is not passed in Right now, I'm tweaking a function that has a bunch of optional parameters. I would like to be able to set a default value for the very last one if it is not passed in. This essentially looks like: if (zend_parse_parameters(ZEND_NUM_ARGS() TSRMLS_CC, "r|ssl", &link, &arg1, &arg1_len, &arg2, &arg2_len, &long, &long_len) == FAILURE { RETURN_FALSE; } if (!long || long==NULL) { long=DEFAULT_VALUE; } However, what I found when printing out the value of "long" is that it has been set to 1?! I imagine this was by the zend_parse_parameters function. Is there a way to disable it from setting values to optional parameters? --Quanah -- Quanah Gibson-Mount Principal Software Developer ITSS/Shared Services Stanford University GnuPG Public Key: http://www.stanford.edu/~quanah/pgp.html -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php