Php, invokeArgs: parameters changed, how to return them?

Question

First of all, I want to test a function:

private function testMe (array &$output)
{
    $output['a'] = 3; // $$$$ $output gets changes
}

I made a little method to make it public, and call:

public static function makePublicAndCall ($objectInstance, $methodname)
{
    $ref = new ReflectionMethod (get_class($objectInstance), $methodname);
    $ref->setAccessible(true);

    $params = array();
    for ($i = 2+1; $i <= func_num_args(); $i++)
    {
        $params[] = func_get_arg($i-1);
    }
    $result = $ref->invokeArgs ($objectInstance, $params);
    for ($i = 2+1; $i <= func_num_args(); $i++)
    {
        // write back $$$$ here I would need something like "func_get_arg($i-1)"
    }
    return $result;
}

so, using it:

$output = array();
::makePublicAndCall ($object, 'testMe', $output);
// $output OMG output remains the same! It must not be empty but [a] => 3

see the problem? This method has 2 obligatory parameters, and all others are optional (they go to the invoked method itself). But if those parameters are changed, cannot be carried back!

bwoebi · Accepted Answer · 2014-09-29 16:46:40Z

2

For PHP 5.6 and above

PHP 5.6 introduced variadic arguments, which can also accept parameters by reference.

function makePublicAndCall ($objectInstance, $methodname, &...$args) { }

now just forward the $args array filled with arguments by ref to $objectInstance->$methodname

function makePublicAndCall ($objectInstance, $methodname, &...$args) {
    $ref = new ReflectionMethod (get_class($objectInstance), $methodname);
    $ref->setAccessible(true);
    return $ref->invokeArgs($objectInstance, $args);
}

makePublicAndCall($object, 'testMe', $output);

For PHP 5.4 and 5.5 versions

No, way, sorry.

For PHP 5.3 and below

Call-time pass by reference still works with these archaic versions, so feel free to use it.

function makePublicAndCall ($objectInstance, $methodname) {
    $ref = new ReflectionMethod (get_class($objectInstance), $methodname);
    $ref->setAccessible(true);
    return $ref->invokeArgs ($objectInstance, $args);
}
@makePublicAndCall($object, 'testMe', &$output); // note the & here...

Also, you don't have to expect a reference in your testMe function, you get an array filled with references, that's enough; you don't need to get an array filled with references by ref to manipulate the references.

edited Sep 29, 2014 at 16:46

answered Sep 29, 2014 at 1:04

bwoebi

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1 Comment

John Smith Over a year ago

somehow the "For all PHP 5+ versions" version doesnt work... I have to stuck with Php 5.3

John Smith · Accepted Answer · 2014-10-10 00:40:00Z

0

I solved it, regardless of versions:

public static function makePublicAndCall ($objectInstance, $methodname, array &$params = array())
{
    $ref = new ReflectionMethod (get_class($objectInstance), $methodname);
    $ref->setAccessible(true);

    return $ref->invokeArgs ($objectInstance, array(&$params));
}

calling:

::makePublicAndCall ($object, 'testMe', $output);

EDIT: another solution regardless of version, but I bet you will throw your hat to the floor:

public static function makeStaticMethodPublicAndCall ($className, $method, &$params1 = null, &$params2 = null, &$params3 = null, &$params4 = null)
{
    $className = new ReflectionClass($className);
    $method = $className->getMethod($method);
    $method->setAccessible(true);
    return $method->invokeArgs (null, array(&$params1, &$params2, &$params3, &$params4));
}

    .
    .
    .

::makeStaticMethodPublicAndCall ('MyClass', 'myMethod', $param1, $param2);

edited Oct 10, 2014 at 0:40

answered Sep 29, 2014 at 11:01

John Smith

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2 Comments

bwoebi Over a year ago

That wasn't exactly what you asked. That's passing an array by ref, not variadic arguments...

John Smith Over a year ago

yeah youre right. First it did solved my problems, but I realized its not good enough

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