Question

I’m getting this error:
“PHP Parse error: syntax error, unexpected T_VARIABLE in /var/www/vhosts/… on line 66”

Here’s my code:

function combine($charArr, $k) {

    $currentsize = sizeof($charArr);
    static $combs = array();
    static $originalsize = $currentsize; ###### <-- LINE 66 ######
    static $firstcall = true;

    if ($originalsize >= $k) {

        # Get the First Combination 
        $comb = '';
        if ($firstcall) { //if this is first call
            for ($i = $originalsize-$k; $i < $originalsize; $i++) {
                $comb .= $charArr[$i];
            }
            $combs[] = $comb; //append the first combo to the output array
            $firstcall = false; //we only want to do this during the first iteration
        }
    ....
    ....
}

Any idea what’s wrong?

Answer 1

Quoting the manual (that page is about static properties, but the same applies for variables) :

Like any other PHP static variable, static properties may only be
initialized using a literal or
constant; expressions are not
allowed. So while you may initialize
a static property to an integer or
array (for instance), you may not
initialize it to another variable, to
a function return value, or to an
object.

You are using this :

static $originalsize = $currentsize;

Which is initializing with an expression — and not a constant.

And here’s [the manual’s section][2] that says quite the same about static variables :

Static variables may be declared as
seen in the examples above. Trying to
assign values to these variables which
are the result of expressions will
cause a parse error.

And, just in case, here’s about expressions.

In your case, to avoid that problem, I suppose you could modify your code, so it looks like this :

$currentsize = sizeof($charArr);
static $originalsize = null;
if ($originalsize === null) {
    $originalsize = $currentsize;
}

With that :

• The static variable is initialized with a constant
• If its value is the constant one, assign the dynamic value.