Static variable for optimization

Question

I'm wondering if I can use a static variable for optimization:

public function Bar() {
    static $i = moderatelyExpensiveFunctionCall();
    if ($i) {
        return something();
    } else {
        return somethingElse();
    }
}

I know that once $i is initialized, it won't be changed by by that line of code on successive calls to Bar(). I assume this means that moderatelyExpensiveFunctionCall() won't be evaluated every time I call, but I'd like to know for certain.

Once PHP sees a static variable that has been initialized, does it skip over that line of code? In other words, is this going to optimize my execution time if I make a lot of calls to Bar(), or am I wasting my time?

Answer 1

I find it easier to do something like the code below. That way the caching is done globally instead of per implementation of the function.

function moderatelyExpensiveFunctionCall()
{
   static $output = NULL;
   if( is_null( $output ) ) {
     //set $output
   }
   return $output;
}

Answer 2

static $i = blah() won't compile, because php doesn't allow expressions and function calls in static initializers. You need something like

function foo() {
   static $cache = null;

   if(is_null($cache)) $cache = expensive_func();

   do something with $cache
}

Answer 3

This should work in your (quite simple) case:

function your_function() {
    static $output;

    if (!isset($output)) {
       $output = 'A very expensive operation';
    }

    return $output;
}

As for a global caching mechanism, you may use a method similar to this one.

Answer 4

Here is quite shorter approach:

function stuff()
{
    static $smthg = []; // or null, or false, or something else
    if ($smthg) {
        return $smthg;
    }

    // filling $smthg goes here 
    // with a lot of 
    // code strings

    return $smthg;
}

Answer 5

How about:

if (!isset($i))
{
    static $i = moderatelyExpensiveFunctionCall();
}