Implementing interfaces for int long etc.

Question

5.00/5 (2 votes)

See more:

Say I have the following :

C#

public class SomeClass<T> : where T : ICompareable<T> , IGetBytes<T>
{...

Now when I use the following :

C#

var s = new SomeClass<int>();

I get an error that :

The type 'int' cannot be used as type parameter 'T' in the generic type or method 'SomeClass<T>'. There is no boxing conversion from 'int' to 'IGetBytes<int>'.

How can I implement the IGetBytes for an int etc.

Posted 9-Nov-11 20:28pm

Mehdi Gholam

Updated 9-Nov-11 20:52pm

Sander Rossel

v2

Add a Solution

2 solutions

Solution 2

What you are saying is that the type of T should be an IComparable of itself (int is an IComparable<int>) and that type T should also be an IGetBytes of <T>.
Int32 on MSDN[^] shows that the int structure is no IGetBytes<int> so it can not be used as type T here.
Look at Shmuel Zangs answer for a workaround. Although I am not sure you should use MyWrapperClass<T> because the implementation of IComparable and IGetBytes probably won't be generic for all types... Rather just create a MyIntWrapper Class.

Posted 9-Nov-11 20:57pm

Sander Rossel

Updated 9-Nov-11 20:58pm

v2

Comments

Mehdi Gholam 10-Nov-11 3:02am

There should be a way with say extension methods etc. ?

Actually I have done a similar thing with byte[] which I have mapped to a bytearr class of my own, but it seems a bit too much and there should be an easier way than reimplementing base types.

Sander Rossel 10-Nov-11 3:46am

As far as I know Extension Methods only extend with Methods (kind of as the name implies). Extending the behaviour of a complete type is usually done through Inheritance or composition... It would be interesting if this was possible though.

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Shmuel Zang · Accepted Answer · 2011-11-09T20:52:00

You cannot implement interfaces in int because the code of System.Int32 isn't yours. But you can create a wrapper class that implements the interfaces as the following:

C#

public class MyWrapperClass<T> : ICompareable<T>, IGetBytes<T>
{
    public MyWrapperClass(T arg)
    {
        // ...
    }

    // ...
}

and use your class with the wrapper:

C#

public class SomeClass<T>
{
    void MyMethod(T arg)
    {
        MyWrapperClass<T> wrappedArg = new MyWrapperClass<T>(arg);

        // ...
    }
}