Checking if an instance's class implements an interface?

Given a class instance, is it possible to determine if it implements a particular interface? As far as I know, there isn't a built-in function to do this directly. What options do I have (if any)?

interface IInterface
{
}

class TheClass implements IInterface
{
}

$cls = new TheClass();
if ($cls instanceof IInterface) {
    echo "yes";
}

You can use the "instanceof" operator. To use it, the left operand is a class instance and the right operand is an interface. It returns true if the object implements a particular interface.

As therefromhere points out, you can use class_implements(). Just as with Reflection, this allows you to specify the class name as a string and doesn't require an instance of the class:

interface IInterface
{
}

class TheClass implements IInterface
{
}

$interfaces = class_implements('TheClass');

if (isset($interfaces['IInterface'])) {
    echo "Yes!";
}

class_implements() is part of the SPL extension.

See: http://php.net/manual/en/function.class-implements.php

Performance Tests

Some simple performance tests show the costs of each approach:

Given an instance of an object

Object construction outside the loop (100,000 iterations)





class_implements
Reflection
instanceOf




140 ms
290 ms
35 ms


'--------------------------------------------'





Object construction inside the loop (100,000 iterations)




class_implements
Reflection
instanceOf




182 ms
340 ms
83 ms


431 ms
607 ms
338 ms


'--------------------------------------------'










Given only a class name
100,000 iterations





class_implements
Reflection
instanceOf




149 ms
295 ms
N/A


'--------------------------------------------'










Where the expensive __construct() is:

public function __construct() {
    $tmp = array(
        'foo' => 'bar',
        'this' => 'that'
    );  

    $in = in_array('those', $tmp);
}

These tests are based on this simple code.

nlaq points out that instanceof can be used to test if the object is an instance of a class that implements an interface.

But instanceof doesn't distinguish between a class type and an interface. You don't know if the object is a class that happens to be called IInterface.

You can also use the reflection API in PHP to test this more specifically:

$class = new ReflectionClass('TheClass');
if ($class->implementsInterface('IInterface'))
{
  print "Yep!\n";
}

See http://php.net/manual/en/book.reflection.php

Just to help future searches is_subclass_of is also a good variant (for PHP 5.3.7+):

if (is_subclass_of($my_class_instance, 'ISomeInterfaceName')){
    echo 'I can do it!';
}

Update

The is_a function is missing here as alternative.

I did some performance tests to check which of the stated ways is the most performant.

Results over 100k iterations

      instanceof [object] took   7.67 ms | +  0% | ..........
            is_a [object] took  12.30 ms | + 60% | ................
             is_a [class] took  17.43 ms | +127% | ......................
class_implements [object] took  28.37 ms | +270% | ....................................
       reflection [class] took  34.17 ms | +346% | ............................................

Added some dots to actually "feel" see the difference.

Generated by this: https://3v4l.org/8Cog7

Conclusion

In case you have an object to check, use instanceof like mentioned in the accepted answer.

In case you have a class to check, use is_a.

Bonus

Given the case you want to instantiate a class based on an interface you require it to have, it is more preformant to use is_a. There is only one exception - when the constructor is empty.

Example: is_a(<className>, <interfaceName>, true);

It will return bool. The third parameter "allow_string" allows it to check class names without instantiating the class.

You can also do the following

public function yourMethod(YourInterface $objectSupposedToBeImplementing) {
   //.....
}

It will throw an recoverable error if the $objectSupposedToBeImplementing does not implement YourInterface Interface.