CODEWITHSUNDEEP
c#
santosh singh
santosh singh

Reputation: 28672

Static field initializers in c#

I have following code in c#

class Test
{
  public static int X = Y;    
  public static int Y = 3;    
}

static void Main()
{
Console.WriteLine(Test.X);
Console.WriteLine(Test.Y);
}

In this case I am getting 0 and 3 but in the following case I am getting 3,3 

class Test
{
  public static int X = 3;    
  public static int Y = X;    
}

static void Main()
{
Console.WriteLine(Test.X);
Console.WriteLine(Test.Y);
}

why is it so?

Upvotes: 9

Views: 821

Answers (6)

BoltClock
BoltClock

Reputation: 724392

From the C# spec:

The static field variable initializers of a class correspond to a sequence of assignments that are executed in the textual order in which they appear in the class declaration.

Therefore, in your first snippet, since Y isn't initialized yet when X is, X has to default to 0 (because default(int) == 0). On the other hand, in your second snippet, Y can be initialized to the value of X because X has already been given a value of 3 in the immediately-preceding statement.

Upvotes: 16

Tipx
Tipx

Reputation: 7515

Correct me if I'm wrong (or downrate me, w/e) but ints are value type in C#. When you first assign X to Y's value, Y is worth 0, and it copies the value, and not the reference.

Upvotes: 1

Shekhar_Pro
Shekhar_Pro

Reputation: 18430

Simply because the X is declared before Y and is also Initialized in that order. So by default an int is 0 so X was Zero when Y was not initialized.

Upvotes: 1

Chris Barlow
Chris Barlow

Reputation: 3314

Geek -

When you instantiate an int, it's default value is 0. Hence, when you run:

public static int X = Y;    

public static int Y = 3;

...the Y is "0" when you set X equal to it, and you are setting Y to 3. But when you run:

public static int X = 3;    

public static int Y = X;

You are setting X equal to 3 and THEN setting Y = X (which is now 3).

Upvotes: 4

Thomas Levesque
Thomas Levesque

Reputation: 292685

Fields are initialized in the order of their declaration. In the first case, you initialize X with the value of Y, but Y isn't initialized yet, so its value is 0. In the second case, you initialize Y with the value of X, which is 3, so both fields contain 3.

Upvotes: 1

jason
jason

Reputation: 241769

This is per the specification, which states the order for which static fields are initialized. The basic point is that they are initialized in the order that they are declared.

Consequently, in your first snippet, X is initialized first and then Y. As Y has not been initialized yet, it has the default value of 0 and so X gets the value 0.

In your second snippet, X is initialized first but is given the explicit value of 3. Then, Y is initialized and is given the value of X which is 3 since it was initialized first.

From §10.5.5.1 Static field initialization

The static field variable initializers of a class correspond to a sequence of assignments that are executed in the textual order in which they appear in the class declaration. If a static constructor (§10.12) exists in the class, execution of the static field initializers occurs immediately prior to executing that static constructor.

Upvotes: 7

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