Can I bind one singleton to two interfaces normally?

I want to do something like that: bind(IFoo1.class).to(Foo.class).singletonInScope() bind(IFoo2.class).to(Foo.class).singletonInScope() but in that case will be created two instances of Foo. So are there any ways to bind Foo class to IFoo1 and IFoo2 interfaces by creating only one instance of Foo. And the other requirement - to initialize Foo.class automatically.

Thank you!

[stephanenicolas]

Toothpick doesn’t support aliases. But in this case you can do:

Foo foo = new Foo();
bind(IFoo1.class).toInstance(foo);
bind(IFoo2.class).toInstance(foo);

Re-open if needed.

[dlemures]

@Dimchel @morder @dimsuz, there are 2 reason why we didn’t implement it:

1. On TP we are trying to keep it simple. Having such a feature, will allow to do something like the following code. It can be difficult to follow (error prone) and affect TP speed:

A -> B
B -> C
D -> B

1. The way TP is implemented, bindings and scope annotations somehow provide a similar information: where and how they are created and for how long they are alive. (not exactly your scenario, but very similar)

However, it is possible to achieve the same by doing it more explicit: using a Provider.

class MyProvider<T>(val scope: Scope, val cls: Class<T>) : Provider<T> {

    override fun get(): T {
        return scope.getInstance(cls)
    }
}

Then you can use it:

bind(Foo.class).singletonInScope() 
bind(IFoo1.class).toProviderInstance(MyProvider(scope, Foo.class))
bind(IFoo2.class).toProviderInstance(MyProvider(scope, Foo.class))

This sample might not be optimal, but it is to show the idea.