Value doesn't change when using a struct in a class

Question

Recently, I found a post on Internet asking this question, I tried to figure out but not sure.

I guess the problem is related with Boxing and Unboxing:

public readonly object counter = new Counter();

This line boxs a Counter onto heap, and counter refers to it.

((Counter)riddle.counter)

This line unboxs the Counter from heap.

Every time data it unboxs from heap is same as origin. Therefore, Line A doesn't affect Line B because they both retrieve from heap and are two different instances of Counter.

Is that right? Sorry for my poor English.

public void WantToKnow()
{
    var riddle = new Riddle();
    ((Counter)riddle.counter).Increment();  // Line A

    Console.WriteLine(((Counter)riddle.counter).Count); // Line B
    // Why the output is 0?///////////////////
}

struct Counter
{
    private int x;
    public void Increment() { this.x++; }
    public int Count { get { return this.x; } }
}

class Riddle
{
    public readonly object counter = new Counter();
}

Remove the readonly from counter variable inside the Riddle class. — Raviraj Palvankar
– Raviraj Palvankar, Commented May 7, 2018 at 6:04
@RavirajPalvankar: That wouldn't change anything in itself. The OP could rebox the modified counter and assign it back to the field, but that's different. — Jon Skeet
– Jon Skeet, Commented May 7, 2018 at 6:49
@DaisyShipton Remove the readonly and also change the datatype from object to Counter and remove all the type casting done inside WantToKnow(), it will increment the counter. The typecasting is treating it as a new object. — Raviraj Palvankar
– Raviraj Palvankar, Commented May 7, 2018 at 6:53

Jon Skeet · Accepted Answer · 2018-05-07 06:49:43Z

Firstly, this is all a good example of why I try to avoid mutable structs in almost all cases. While you can usually predict what will happen if you pay enough attention, it's easy to miss one copy along the way which messes everything up. If you keep all structs immutable, life is simpler.

For your question: yes, you're unboxing which creates a copy of the counter. Changing that copy doesn't affect anything else. (There's IL to unbox without copying, but C# never uses that.)

You can make this work by making your counter implement an interface with an Increment operation. At that point you can cast to the interface instead of the value type, which means it's not unboxing. Your Increment operation would then modify the value within the box, which means you can then get at it again. Here's a complete example:

using System;

class Program
{
    static void Main()
    {
        var riddle = new Riddle();
        ((ICounter)riddle.counter).Increment();

        Console.WriteLine(((Counter)riddle.counter).Count); // Line B
    }
}

interface ICounter
{
    void Increment();
}

struct Counter : ICounter
{
    private int x;
    public void Increment() { this.x++; }
    public int Count { get { return this.x; } }
}

class Riddle
{
    public readonly object counter = new Counter();
}

Siva Rm K · Accepted Answer · 2018-05-07 06:50:15Z

0

Because struct is value type- You can use "class Counter instead "struct Counter" or use below solution
Reason : ((Counter)riddle.counter).Count is treat as another copy of struct not as ref type. so it's showing initial value 0

using System;
using System.Text.RegularExpressions;

public class Program
{
    public static void Main()
    {

         var riddle = new Riddle();
             Console.WriteLine(((Counter)riddle.counter).Increment());
            ///////////////////////////////////////////
            Console.WriteLine(((Counter)riddle.counter).Count);
            // Why the output is 0?///////////////////
    }


}
struct Counter
        {
            private int x;

            //Change is here    
            public int Increment() { this.x++; return this.x; }
            public int Count { get { return this.x; } }
        }

        class Riddle
        {
            public readonly object counter = new Counter();
        }

answered May 7, 2018 at 6:50

Siva Rm K

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3 Comments

Jon Skeet Over a year ago

That wouldn't do very much good though - try copying your line of code that calls Increment several times, and you'll see that it's still not actually implementing the counter in Riddle. The final output is still 0.

Siva Rm K Over a year ago

Yes.I got that.

Jon Skeet Over a year ago

So how is this a "solution"?

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