Returning array inside function that returns another value

Question

Why does the following script does not work?

$arr = array();

function collect( $array , $val) {
    $array[] = $val;
    return $array;
}

function checkFoo( $s ) {

    $valid = true;

    if ( strlen( $s ) === 0 ) {
        $isValid = false;
        collectFoo( $arr , $s );
    }

    return $valid;
}

function checkBar( $t ) {

    $valid = true;

    if ( strlen( $s ) != 10 ) {
        $isValid = false;
        collectFoo( $arr , $t );
    }

    return $valid;
}

if ( checkFoo( $that ) && checkBar( $this ) ) {

    echo "success"; 

} else {

    print_r( $error );

}

I always get

Notice: Undefined variable: error in /my.php on line 12

where line 12 resembles the second occurrence of collect(...);

I know that a function can only return one value, but what if a function returns something in a function that returns something? Because collect returns $array inside checkBar that returns $valid.

what is $this and $that and $error and collectFoo function ? — DJafari
– DJafari, Commented Feb 15, 2013 at 22:52
$this & $that are placeholders. $error is an array that is just still there from debugging, collectFoo is a function that adds something to an array. — Sven
– Sven, Commented Feb 15, 2013 at 23:08

Alon Eitan · Accepted Answer · 2013-02-15 22:49:46Z

1

you are using a global variable ($arr), so you need do declare it as such. example:

function checkFoo( $s ) {
    global $arr; // declare $arr to be usable inside this function scope
    $valid = true;

    if ( strlen( $s ) === 0 ) {
        $isValid = false;
        collectFoo( $arr , $s );
    }

    return $valid;
}

answered Feb 15, 2013 at 22:49

Alon Eitan

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Comments

Community · Accepted Answer · 2017-05-23 12:27:03Z

0

You aren't declaring your $arr variable as global inside the functions that use it. You can find some info about that here.

Also, it doesn't look like you are actually using the return value of collect, so I'd say that your problem is that either checkBar or checkFoo are returning false, which makes you fall through to the print_r function, with a $error variable that you have not initialized.

Finally, it seems to me like your collect function is not actually doing anything to the $arr variable, because you are not passing it to collect by reference. you can read up on passing variables by reference here. There is also a S.O. question about this here.

edited May 23, 2017 at 12:27

CommunityBot

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answered Feb 15, 2013 at 23:04

SuperTron

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Comments

DJafari · Accepted Answer · 2013-02-15 23:11:32Z

0

Edited

You need more practice, because totally your code is incorrect ! you must changes this parts of your codes :

$arr = array();

function collectFoo( $arr , $val) {
    global $arr;

    $arr[] = $val;
}

function checkFoo( $s ) {
    global $arr;

    $valid = false;
    if ( strlen( $s ) === 0 ) {
        $valid = false;
        collectFoo( $arr , $s );
    }
    return $valid;
}

function checkBar( $t ) {
    global $arr;

    $valid = true;
    if ( strlen( $t ) != 10 ) {
        $valid = false;
        collectFoo( $arr , $t );
    }
    return $valid;
}

$a = checkFoo( $that );
$b = checkBar( $this );
if ( $a && $b ) {
    echo 'Success !';
} else {
    print_r( $err );
}

edited Feb 15, 2013 at 23:11

answered Feb 15, 2013 at 23:02

DJafari

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1 Comment

Sven Over a year ago

Sorry, but I only get Array ( [0] => [1] => )

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