Re: Conditional compilation
| From: | Shawn McKenzie | Date: | Fri, 15 Aug 2008 23:34:04 +0000 |
| Subject: | Re: Conditional compilation | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-278205@lists.php.net to get a copy of this message | ||
Herman Gomez wrote:
Hi, Here is something I used to do in C/C++ to include/exclude automaticaly all debugging code at compiling time: #define debug TRUE #ifdef(debug)Well PHP isn't compiled it's interpreted. Still I don't see much diff and no overhead between the following: #ifdef(debug) //debugging code #endif ---and--- if (defined('DEBUG')) { //debugging code } I don't think checking a define is cpu intensive or even measurable. You could "assume" that it's defined as true or false and: if (DEBUG === true)) { //debugging code } Still, I don't think that even checking $debug is measurable. -Shawn//debugging code#endif That way I can include/exclude easily all debugging code in the final compiled code. In PHP I have not been able to find anything like that. The only solution I've found is having this kind of code in every debug code block: if ($debug) {//debugging code} But this means that the debugging code is in the final compiled (interpreted) code, wasting cpu cycles even if there won't be any debugging in production. Does somebody know if there is something like conditional compilation in PHP that I can use? Regards, Herman Gomez Madrid, Spain.