Not able to print variables?

[hearn]

I'm using the following function:

function do_math($a, $b, $delta_x) {
	$sum = 0;
	$n_counter = 0;

do {
	$sum+=(do_function($a)*$delta_x);
	$a+=$delta_x;	
	$n_counter++;
	$sum_store[$n_counter] = $sum;
} while($a<$b);

return($sum);
}

My problem is that I'm not able to reference the variable sum_store from outside the function, it won't print. Does the variable or function need to be globalized somehow for this to work?

thanks in advance,
Cheers!

[gtibok]

hi

I think your function should look like this:
<?php
function do_math($a, $b, $delta_x) {
$sum = 0;
$n_counter = 0;
do {
$sum+=$a*$delta_x;
$a+=$delta_x;

$n_counter++;
$sum_store[$n_counter] = $sum;
} while($a<$b);
return($sum_store[$n_counter]);
}
echo do_math($a = 1, $b = 9, $delta_x = 3);
?>

but what it's this about? ... I don't understand it's meaning

[archangel_617b]

I believe you need to double check your return statement. If you want to refer to $sum_store outside the function, return that instead of $sum 😛

Now if $sum_store is a variable outside the function, you ought to pass it in by reference like so:

function do_math($a, $b, $delta_x, &$sum_store) 
{
     //  ... your beautiful code in here 
     return $sum;
}

In this way, any changes to $sum_store in the function will affect the variable outside of it. I recommend you do this instead of making it global since that leads to problems if you call do_math() from within another function.

Is clear? If not, the documentation on "pass-by-reference" should answer many more of your questions.