Php updating a static variable internet dating movie

Posted by / 27-Aug-2019 18:55

Creating new lambda with function() ; expression creates new instance of its static variables.

Assigning a lambda to a variable does not create a new instance.

I guess PHP keeps a cached "template" for the anonymous function and therefore doesn't need much time to create a new instance of the closure! It's a pretty common pattern in some other famous scripting languages.

And I tried 10 million loop iterations, in PHP 7.0.14 from Dec 2016.

Result:a single saved closure kept in a variable and re-used (10000000 iterations): 1.3874590396881 secondsnew anonymous closure created each time (10000000 iterations): 2.8460240364075 seconds In other words, over the course of 10 million iterations, creating the closure again during every iteration only added a total of "1.459 seconds" to the runtime.

At least, it will be if you don't reassign $fib to anything else between creating the function and calling it: Of course, that's true of any variable: if you don't want its value to change, don't change its value.

All the usual scoping rules for variables still apply: a local variable in a function is a different variable from another one with the same name in another function: If you want to make a recursive closure, you will need to write this:$some_var1="1";$some_var2="2";function($param1, $param2) use ($some_var1, $some_var2)If you need to pass values by reference you should check out you're wondering if $some_var1 and $some_var2 are still visible by using the call_user_func, yes, they are available.

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Consider the following example: That means, regular method invocations work like expected and like before.

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