readonly (C# Reference)


In this example, the value of the field year cannot be changed in the method ChangeYear, even though it is assigned a value in the class constructor:

    class Age
        readonly int _year;
        Age(int year)
            _year = year;
        void ChangeYear()
            //_year = 1967; // Compile error if uncommented.

You can assign a value to a readonly field only in the following contexts:

  • When the variable is initialized in the declaration, for example:

    public readonly int y = 5;
  • For an instance field, in the instance constructors of the class that contains the field declaration, or for a static field, in the static constructor of the class that contains the field declaration. These are also the only contexts in which it is valid to pass a readonly field as an out or ref parameter.

Note Note

The readonly keyword is different from the const keyword. A const field can only be initialized at the declaration of the field. A readonly field can be initialized either at the declaration or in a constructor. Therefore, readonly fields can have different values depending on the constructor used. Also, while a const field is a compile-time constant, the readonly field can be used for runtime constants as in the following example:

public static readonly uint timeStamp = (uint)DateTime.Now.Ticks;
    public class ReadOnlyTest
       class SampleClass
          public int x;
          // Initialize a readonly field 
          public readonly int y = 25;
          public readonly int z;

          public SampleClass()
             // Initialize a readonly instance field
             z = 24;

          public SampleClass(int p1, int p2, int p3)
             x = p1;
             y = p2;
             z = p3;

       static void Main()
          SampleClass p1 = new SampleClass(11, 21, 32);   // OK
          Console.WriteLine("p1: x={0}, y={1}, z={2}", p1.x, p1.y, p1.z);
          SampleClass p2 = new SampleClass();
          p2.x = 55;   // OK
          Console.WriteLine("p2: x={0}, y={1}, z={2}", p2.x, p2.y, p2.z);
        p1: x=11, y=21, z=32
        p2: x=55, y=25, z=24

In the preceding example, if you use a statement like this:

p2.y = 66; // Error

you will get the compiler error message:

The left-hand side of an assignment must be an l-value

which is the same error you get when you attempt to assign a value to a constant.
