I'm getting a “syntax error, unexpected T_VARIABLE” error. I don't see what I'm doing wrong?

问题

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?

回答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 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.

回答2:

static $originalsize = $currentsize; ###### <-- LINE 66 ######

You can't pass a variable as the default value of a static variable. Instead, do the following:

static $originalsize;
$originalsize = $currentsize;

回答3:

To quote the php manual:

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.

回答4:

From php manual:

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.

来源：https://stackoverflow.com/questions/5122729/im-getting-a-syntax-error-unexpected-t-variable-error-i-dont-see-what-im

标签

php

syntax-error

parse-error