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Home/ Questions/Q 801439
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T23:26:29+00:00 2026-05-14T23:26:29+00:00

So according to my testing, If you have something like: Module modA = new

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So according to my testing, If you have something like:

Module modA = new AbstractModule() {
    public void configure() {
        bind(A.class).to(AImpl.class);
        bind(C.class).to(ACImpl.class);
        bind(E.class).to(EImpl.class);
    }
}

Module modB = New AbstractModule() {
    public void configure() {
        bind(A.class).to(C.class);
        bind(D.class).to(DImpl.class);
    }
}

Guice.createInjector(Modules.overrides(modA, modB)); // gives me binding for A, C, E AND D with A overridden to A->C.

But what if you want to remove the binding for E in modB? I can’t seem to find a way to do this without having to break the bind for E into a separate module. Is there a way?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-14T23:26:29+00:00Added an answer on May 14, 2026 at 11:26 pm

    The SPI can do this. Use Elements.getElements(modA, modB) to get a list of Element objects, representing your bindings. Iterate through that list and remove the binding whose key you’d like to remove. Then create a module from the filtered elements using Elements.getModule(). Putting it all together:

    public Module subtractBinding(Module module, Key<?> toSubtract) {
      List<Element> elements = Elements.getElements(module);
    
      for (Iterator<Element> i = elements.iterator(); i.hasNext(); ) {
        Element element = i.next();
        boolean remove = element.acceptVisitor(new DefaultElementVisitor<Boolean>() { 
          @Override public <T> Boolean visit(Binding<T> binding) { 
            return binding.getKey().equals(toSubtract);
          }
          @Override public Boolean visitOther(Element other) {
            return false;
          }
        }); 
        if (remove) {
          i.remove();
        }
      }
    
      return Elements.getModule(elements);
    }
    
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