Lets say I create and interface
interface IMyInterface {
function abstractMethod();
}
class MyClass implements IMyInterface {
function abstractMethod() {
//code
}
}
class OtherClass {
private $IMyInterfaceObj;
function __construct($obj) {
$this->IMyInterfaceObj = $obj;
}
}
What can I do to make sure that the object assigned to $IMyInterfaceObj is an Object that actually implements the interface, since PHP is loosely typed. Should I check the type???
You would type hint it in the constructor. You cannot do this for basic types such as integers or strings, although you can for arrays with array. The only value you can use to make a parameter an optional one is to use null.
class OtherClass {
private $IMyInterfaceObj;
function __construct(IMyInterface $obj) {
$this->IMyInterfaceObj = $obj;
}
}
Thoroughly reading the documentation on interfaces and type hinting should clear anything else up on the subject.