CODEWITHSUNDEEP
php
John Smith
John Smith

Reputation: 233

Storing a callback function in a variable gives an error

I am studying the book "Programming PHP 3rd edition", and I came upon these lines go code:

$callback = function myCallbackFunction()
{
echo "callback achieved";

} call_user_func($callback);

But, when I try to run it, I get the following error:

syntax error, unexpected 'myCallbackFunction' (T_STRING), expecting'('

Am I doing something wrong? Is the code incorrect?
And here is my PHP version:

% php -v                                                                                                                  255 ↵
PHP 5.4.19 (cli) (built: Sep 18 2013 14:26:26)
Copyright (c) 1997-2013 The PHP Group
Zend Engine v2.4.0, Copyright (c) 1998-2013 Zend Technologies
    with XCache v2.0.1, Copyright (c) 2005-2012, by mOo
    with the ionCube PHP Loader v4.5.2, Copyright (c) 2002-2014, by ionCube Ltd.

Upvotes: 0

Views: 62

Answers (3)

Daniel P
Daniel P

Reputation: 2449

Looks like you're trying to create an anonymous function.

$callback = function()
{
    echo "callback achieved";
}

call_user_func($callback);

Try removing the functions name.

If you are trying to call an existing function, then you can do it like this:

function myCallbackFunction()
{
    echo "callback achieved";
}

call_user_func('myCallbackFunction');

Upvotes: 0

hek2mgl
hek2mgl

Reputation: 158010

Anonymous functions don't have a function name. That's why they are called anonymous. The definition should look like this:

$callback = function ()
{
    echo "callback achieved";
};

call_user_func($callback);

Upvotes: 2

idmean
idmean

Reputation: 14875

Either you define an anonymous function or a non-anonymous.

To define a anoymous, what are you obivously trying to do, you shouldn't put a name before the argument list ()

$callback = function()
{
echo "callback achieved";

};
call_user_func($callback);

Upvotes: 0

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