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one interface implementing another

I want to have the two interfaces below.

public interface IWriteable<T>
                where T : class
{
    void Add(T newEntity);
    void Remove(T entity);
}

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public interface IReadable<T>
            where T : class
{
    IQueryable<T> FindAll();
    IQueryable<T> FindWhere(Expression<Func<T, bool>> predicate);
    T FindById(int id);
    T FindById(Guid id);
}

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Basically I want to control certain elements of different aspects of my website.

So some repositories will only be able to read certain classes, some read/write.

What I want though is for IWriteable to also Implement IReadable as my attitude is if something has the permission to write it is ok to read.

Using my interface above how to I make Iwritebale implement IReadabale?
0
scm0sml
Asked:
scm0sml
  • 2
1 Solution
 
käµfm³d 👽Commented:
Interfaces cannot provide implementation details, but you can certainly mark you IWriteable as "implementing" (only word I could think of, sorry) the IReadable interface, whose implementation responsibilities would also be passed along to classes implementing IWriteable. You simply need to note it in the implementation list (again, sorry). For example:

public interface IWriteable<T> : IReadable<T>
                where T : class
...

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0
 
scm0smlAuthor Commented:
Thats exactly what I was after thanks!

Probably didint explain myself very well!
0
 
käµfm³d 👽Commented:
Probably didint explain myself very well!
I'd say you explained it just fine  = )

Glad to help!
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