Checking if an instance's class implements an interface?

Question

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 opt

Answer 1

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      | Cheap Constructor
| 431 ms           | 607 ms     | 338 ms     | Expensive Constructor
'--------------------------------------------'

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.