static variable allocation time in .net

Question

When static variable is allocated i.e. when you declare class or at the time of object creation?

See this related question: stackoverflow.com/questions/218461/… — Dirk Vollmar
– Dirk Vollmar, Commented Dec 21, 2009 at 15:31

Teun D · Accepted Answer · 2009-12-21 16:25:27Z

6

It is compiled into the static constructor. So the first time anyone creates an object of the class or calls a static method or property on it, the initialization occurs.

Edit: when it is important to you if initialization happens before of after your own static constructor code (and some other edge cases),check the link in the comment by divo.

edited Dec 21, 2009 at 16:25

answered Dec 21, 2009 at 15:18

Teun D

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4 Comments

Dirk Vollmar Over a year ago

It's not as simple as that. Jon Skeet has a comprehensive article on that, see yoda.arachsys.com/csharp/beforefieldinit.html. In classes without static constructors type initializers might be run when the assembly is loaded or when a static method on the type is executed or even only when the first field of the type is accessed.

Teun D Over a year ago

That is a really interesting link, divo. Thanks. I think that in most practical purposes, my answers stands, but there appears to be more to it than I imagined.

Dirk Vollmar Over a year ago

Yes, as Jon wrote: "Not many classes actually need the behaviour assumed by many C# programmers - most people need never know the difference, really." But it might be exactly these edge cases that leed to unexpected errors.

Teun D Over a year ago

Yeah, the singleton pattern in his samples is one of those, I guess.

Peter · Accepted Answer · 2009-12-21 15:17:33Z

4

the first time the class is accessed....

answered Dec 21, 2009 at 15:17

Peter

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Comments

Adam Ralph · Accepted Answer · 2009-12-22 07:16:20Z

2

As other answers have indicated, this will happen in the type (static) constuctor. If your class does not have a type constructor explicitly defined then the compiler will generate one for you. However, determining exactly when this constructor is called is a bit more involved.

If your class does not define an explicit type constructor, e.g.

public class Foo
{
    public static int Bar = 1;
}

then the C# compiler will generate a constructor and emit the class definition with the beforefieldinit flag. This will cause the JIT compiler to guarantee that the type constructor is called sometime before a member of the type is first used but this time is non-deterministic, i.e. it is not possible to know exactly when this will happen, and it could be at a much earlier point than when a type member is first used.

If your class declares an explicit type constructor, e.g.

public class Foo
{
    public static int Bar;
    static Foo()
    {
        Bar = 1;
    }
}

then the compiler will emit IL for the class without the beforefieldinit flag. In this case the JIT compiler will call the type constructor at a deterministic time, i.e. immediately before the first type member access.

The former JIT behaviour is known as before-field-init semantics and the latter as precise sematntics. It is important to know the difference between the two since, in some scenarios, they may have a significant performance implication.

edited Dec 22, 2009 at 7:16

answered Dec 21, 2009 at 15:35

Adam Ralph

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2 Comments

LukeH Over a year ago

A nitpick: What matters for beforefieldinit is whether or not a static constructor exists at all, not whether your fields are initialised inside the constructor. If the class has no static constructor then it will be marked as beforefieldinit by the compiler; if the class has a static constructor then it won't be marked as beforefieldinit, regardless of where its fields are initialised.

Adam Ralph Over a year ago

@Luke - thanks for pointing that out. Nitpicking is good ;-) have edited

ironic · Accepted Answer · 2009-12-21 15:19:17Z

0

Static variables are allocated as soon as static (type) constructor is called. This happens when you call any method that reference the type for the first time, before method execution.

answered Dec 21, 2009 at 15:19

ironic

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3 Comments

Dirk Vollmar Over a year ago

This is only true if there is a static type constructor.

ironic Over a year ago

Any explicit fields initialization is compiled-in to the constructor before the constructor body. If there is no declared static constructor compiler will generate one.

Dirk Vollmar Over a year ago

@ironic: No, this is not the case. If there is no static constructor the class will be marked with the beforefieldinit flag meaning the time when the type initializer is executed is not deterministic.

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