#44384 [Opn->Csd]: Static variables not very clearly explained

ID: 44384 Updated by: vrana@php.net Reported By: svevo at svevoromano dot it -Status: Open +Status: Closed Bug Type: Documentation problem Operating System: OSX PHP Version: Irrelevant New Comment: This bug has been fixed in the documentation's XML sources. Since the online and downloadable versions of the documentation need some time to get updated, we would like to ask you to be a bit patient. Thank you for the report, and for helping us make our documentation better. "Now, $a is initialized only in first call of function and every time the test() function is called it will print the value of $a and increment it." Previous Comments: ------------------------------------------------------------------------ [2008-03-09 21:20:13] svevo at svevoromano dot it Description: ------------ Given the following example: <?php function Test() { static $a = 0; echo $a; $a++; } ?> I reckon the explanation is lacking one important point: the fact that the static variable line is executed only on the first call to the function (at least this is what emerged in a discussion I had with some of you in the mailing list). I think this point should be clearly stated because beginners may not know how this works in C or other languages. At the moment the exlanation reads: This function is quite useless since every time it is called it sets $a to 0 and prints "0". The $a++ which increments the variable serves no purpose since as soon as the function exits the $a variable disappears. To make a useful counting function which will not lose track of the current count, the $a variable is declared static: Now, every time the Test() function is called it will print the value of $a and increment it. A non-expert could easily think that this variable gets set to zero on each call to the function. ------------------------------------------------------------------------ -- Edit this bug report at http://bugs.php.net/?id=44384&edit=1