php is_function() to determine if a variable is a function

Question

I was pretty excited to read about anonymous functions in php, which let you declare a variable that is function easier than you could do with create_function. Now I am wondering if I have a function that is passed a variable, how can I check it to determine if it is a function? There is no is_function() function yet, and when I do a var_dump of a variable that is a function::

$func = function(){
    echo 'asdf';
};
var_dump($func);

I get this:

object(Closure)#8 (0) { } 

Any thoughts on how to check if this is a function?

Answer 1

Use is_callable to determine whether a given variable is a function. For example:

$func = function()
{  
    echo 'asdf';  
};

if( is_callable( $func ) )
{
    // Will be true.
}

Answer 2

You can use function_exists to check there is a function with the given name. And to combine that with anonymous functions, try this:

function is_function($f) {
    return (is_string($f) && function_exists($f)) || (is_object($f) && ($f instanceof Closure));
}

Answer 3

If you only want to check whether a variable is an anonymous function, and not a callable string or array, use instanceof.

$func = function()
{  
    echo 'asdf';  
};

if($func instanceof Closure)
{
    // Will be true.
}

Anonymous functions (of the kind that were added in PHP 5.3) are always instances of the Closure class, and every instance of the Closure class is an anonymous function.

There's another type of thing in PHP that could arguably be considered a function, and that's objects that implement the __invoke magic method. If you want to include those (while still excluding strings and arrays), use method_exists($func, '__invoke'). This will still include closures, since closures implement __invoke for consistency.

Answer 4

function is_function($f) {
    return is_callable($f) && !is_string($f);
}