Can dependency injection prevent a circular dependency?

Question

Project#1 has some interfaces and classes that project#2 references.

Now I want to use the implementation of Project#2 in Project#1 but vs.net complains about a circ

Answer 1

You probably could solve this with DI, but you shouldn't.

If I understand correctly, you have something like this:

  + Assembly A           + Assembly B
  |                      |
  +-- Interface IFoo     +-- Class ConcreteFoo : IFoo
  |                                   ^
  +-- Class MyClass -->------->-------|

In other words, you're trying to get MyClass to reference ConcreteFoo, but you can't because assembly B, which ConcreteFoo resides in, already depends on IFoo in A.

This is a design error. If you declare the interface IFoo in Assembly A, but no concrete implementations, then any other interfaces/classes in assembly A should only reference IFoo, never a concrete class that implements it.

There are three ways to eliminate the circular dependency:

1. Make MyClass dependent on IFoo instead of ConcreteFoo. This is probably the best option if you can do it. If the issue is that you need a physical instance of IFoo for use in MyClass and don't know where to get one from, then have it take an IFoo in the constructor - let whoever uses MyClass figure out what IFoo to use.

2. Move the interfaces to their own assembly. This is still a reasonably good practice. Your design will look like this:

     + Assembly App       + Assembly Interfaces      + Assembly Concrete
     |                    |                          |
     |                    +-- Interface IFoo         |
     |                    |                  \       |
     +-- Class MyClass    |                   \------+-- Class ConcreteFoo
     |                    |                          |           ^
     +---- Member Foo ->--------------------->-------------------|
   

3. Move MyClass to its own assembly. Effectively your dependency tree will look the same as in #2 above, but if assembly A is much smaller than B then this would require less effort.

Hope that helps.