Define the same PHP function in many namespaces

Question

In order to create a mock version of a global function for unit testing, I define its mock version in the namespace of the SUT class, as is explained here

Consider a case of global function foo() which is used in namespace Bar and in namespace Baz. If I want to use the same mock version of foo() in both places, it seems I am forced to make two declarations of the mock for foo()

/* bar-namespace-mocks.php */
namespace Bar;
function foo(){
   return 'foo mock';
}

/* baz-namespace-mocks.php */
namespace Baz;
function foo(){
   return 'foo mock';
}

This code does not conform to the DRY principal. It would be preferred to have one declaration of the foo mock

/* foo-mock.php */
function foo(){
   return 'foo mock';
}

and then import into each namespace as needed like the following pseudo code:

/* namespace-mocks.php */
namespace Bar{
  import 'foo-mock.php';
} 
namespace Baz{
  import 'foo-mock.php';
}

Importing using include, e.g.

namespace Baz;
include 'foo-mock.php'

does not cause the mock to be declared in the Baz namespace. Is there any way to declare a function in more than one namespace without having more than one version of it?

Answer 1

If you need to abstract away a native function, then make a contract for it, so that you can use dependency injection for the service it provides:

interface FooInterface
{
    public function get(): string;
}

Then provide a default implementation that uses the native function:

class NativeFoo implements FooInterface
{
    public function get(): string
    {
        return \foo();
    }
}

Then your main class might look something like:

class A
{
    protected FooInterface $foo;

    public function __construct(FooInterface $foo = null)
    {
        $this->foo = $foo ?? new NativeFoo();
    }

    public function whatever()
    {
        $foo = $this->foo->get();
    }
}

So, you've got an optional argument in the constructor, and if you don't provide a value, you'll get the native foo function as a default. You'd just do:

$a = new A();
$a->whatever();

But if you want to override that, you can do:

$foo = new SomeOtherFoo();
$a = new A($foo);
$a->whatever();

Then, in your test, you can build an explicit mock:

class MockFoo implements FooInterface
{
    public function get(): string
    {
        return 'some test value';
    }
}

And pass in instance of that:

$foo = new MockFoo();
$a = new A($foo);
$a->whatever();

Or use the PHPUnit mock builder:

$foo = $this->getMockBuilder(FooInterface::class)->... // whatever

If you want a simpler version, you could just use a callable:

class A
{
    protected $foo;

    public function __construct(callable $foo = null)
    {
        $this->foo = $foo ?? 'foo';
    }

    public function whatever()
    {
        $foo = call_user_func($this->foo);
    }
}


// default usage
$a = new A();
$a->whatever();

// test usage
$a = new A(fn() => 'some other value');
$a->whatever();

Answer 2

While I whole-heartedly agree with Alex Howansky's answer, if you're truly hell-bent on using plain functions, then you can always:

namespace Mocks {
    function foo() {
        return 'foo';
    }
}

namespace Bar {
  function foo() {
      return \Mocks\foo();
  }
}

namespace Baz {
  function foo() {
      return \Mocks\foo();
  }
}

namespace asdf {
    var_dump(\Bar\foo(), \Baz\foo());
}

Output:

string(3) "foo"
string(3) "foo"

Though at this point we're really just on our way to reinventing classes, but worse. At some point you're going to wonder how to namespace a variable, and that is simply not possible, so you'll be rolling this into classes at that point.

Classes, interfaces, and traits/inheritance would be most properly leveraged to solve this problem.