C# - passing parameters by reference to constructor then using them from method

Question

In the following code, I am trying to have a method(Work) from class TestClass change the values of some variables in the main program without having to return them. The variables are passed by reference in the TestClass constructor.

class Program
{
    static void Main(string[] args)
    {
        int a, b, c, d;
        a = 5; b = 10; c = 20; d = 25;
        Console.WriteLine("Main before TestClass: a=" + a + " b=" + b + " c=" + c + " d=" + d);
        TestClass testObj = new TestClass(ref a,ref b,ref c,ref d);
        testObj.Work();
        Console.WriteLine("Main after TestClass: a=" + a + " b=" + b + " c=" + c + " d=" + d);
        Console.ReadLine();
    }
}

public class TestClass
{
    int _a, _b, _c, _d;
    public TestClass(ref int a, ref int b, ref int c, ref int d)
    {
        _a = a; _b = b; _c = c; _d = d;
    }

    public void Work()
    {
        Console.WriteLine("Work before changing: a=" + _a + " b=" + _b + " c=" + _c + " d=" + _d);
        _a = 0; _b = 1; _c = 2; _d = 3;
        Console.WriteLine("Work after changing: a=" + _a + " b=" + _b + " c=" + _c + " d=" + _d);
    }
}

However this code returns :

Main before TestClass: a=5 b=10 c=20 d=25
Work before changing: a=5 b=10 c=20 d=25
Work after changing: a=0 b=1 c=2 d=3
Main after TestClass: a=5 b=10 c=20 d=25

Is there any way to have the method change the values of the variables in the Main program? Thank you!

Answer 1

Not possible. You can't store a reference to an int as a member, and just int a is a copy - changes to the copy don't change the original, as you've seen.

You could wrap that int inside a class. You can store a reference to the class and operate on it then.

public class IntWrapper
{
    public IntWrapper( int val = 0 ) { Value = val; }
    public int Value { get; set; }
}

static void Main(string[] args)
{
    IntWrapper a = new IntWrapper(5);
    TestClass testObj = new TestClass(a);
    testObj.Work();
}

public class TestClass
{
    IntWrapper _a;
    public TestClass(IntWrapper a)
    {
        _a = a;
    }

    public void Work()
    {
        Console.WriteLine("Work before changing: a=" + _a.Value);
        _a.Value = 0;
        Console.WriteLine("Work after changing: a=" + _a.Value);
    }
}

Answer 2

You would be better off creating your own wrappers over Int32 in order for changes to be reflected, because once the values are assigned to fields of the class, they are no longer references, but rather different instances of Int32. Consider the following code:

class Integer {
    public int Value;

    public Integer(int value) {
        Value = value;
    }

    public override string ToString() {
        return Value.ToString();
    }
}

class TestClass {
    Integer _a, _b, _c, _d;

    public TestClass(Integer a, Integer b, Integer c, Integer d) {
        _a = a;
        _b = b;
        _c = c;
        _d = d;
    }

    public void Work() {
        _a.Value = 111;
        _b.Value = 222;
        _c.Value = 333;
        _d.Value = 444;
    }
}

So now you have an Integer -- a wrapper class over Int32. The usage will bring you the results:

Integer a = new Integer(0), b = new Integer(0), c = new Integer(0), d = new Integer(0);
Console.WriteLine("a: {0}, b: {1}, c: {2}, d: {3}", a, b, c, d);
new TestClass(a, b, c, d).Work();
Console.WriteLine("a: {0}, b: {1}, c: {2}, d: {3}", a, b, c, d);

The output is:

a: 0, b: 0, c: 0, d: 0
a: 111, b: 222, c: 333, d: 444

You may also find it useful to read more about classes and structs in C#, eg http://msdn.microsoft.com/en-us/library/ms173109.aspx . (Int32 is a struct, whereas in your case you probably need a class)

Answer 3

What you are trying to do can't be done as it wasn't designed to be used in that way. You would have to have the parameters going into the Work method directly instead eg

testObj.Work(ref a, ref b, ref c, ref d);

Answer 4

This behaviour is caused by:

int _a, _b, _c, _d;
public TestClass(ref int a, ref int b, ref int c, ref int d)
{
    _a = a; _b = b; _c = c; _d = d;
}

int is a structure (a value type). This means that you're NOT passing the reference (or pointer) to the fields as you may expect, even if you use the ref keyword. What you're doing instead is assigning to the fields the integer values given as input, nothing more.
To achieve what you're expecting you have to use a reference type (wrap your integer into a class) or reference your integers (through the ref keyword) in the method where you actually change their values:

public void Work(ref int a, ref int b, ref int c, ref int d)
{
   Console.WriteLine("Work before changing: a=" + a + " b=" + b + " c=" + c + " d=" + _d);
   a = 0; b = 1; c = 2; d = 3;
   Console.WriteLine("Work after changing: a=" + a + " b=" + b + " c=" + c + " d=" + _d);
}

The effect of passing by reference is that any change to the parameter in the called method is reflected in the calling method.

To understand more about the ref keyword give a look to: ref (C# Reference) .