Is there a way?
I need all types that implement a specific interface to have a parameterless constructor, can it be done?
I am developing the base code for other developers in my company to use in a specific project.
There's a proccess which will create instances of types (in different threads) that perform certain tasks, and I need those types to follow a specific contract (ergo, the interface).
The interface will be internal to the assembly
If you have a suggestion for this scenario without interfaces, I'll gladly take it into consideration…
Best Solution
Juan Manuel said:
It's an indirect mechanism. The generic allows you to "cheat" and send type information along with the interface. The critical thing to remember here is that the constraint isn't on the interface that you are working with directly. It's not a constraint on the interface itself, but on some other type that will "ride along" on the interface. This is the best explanation I can offer, I'm afraid.
By way of illustration of this fact, I'll point out a hole that I have noticed in aku's code. It's possible to write a class that would compile fine but fail at runtime when you try to instantiate it:
Something derives from ITest<T>, but implements no parameterless constructor. It will compile fine, because String does implement a parameterless constructor. Again, the constraint is on T, and therefore String, rather than ITest or Something. Since the constraint on T is satisfied, this will compile. But it will fail at runtime.
To prevent some instances of this problem, you need to add another constraint to T, as below:
Note the new constraint: T : ITest<T>. This constraint specifies that what you pass into the argument parameter of ITest<T> must also derive from ITest<T>.
Even so this will not prevent all cases of the hole. The code below will compile fine, because A has a parameterless constructor. But since B's parameterless constructor is private, instantiating B with your process will fail at runtime.