C# Lambdas and “this” variable scope

Question

I am wondering whether I can use the this keyword inside a C# lambda, although actually I know that I can but I want to make sure that this isn\'t a bad th

Answer 1

There is nothing wrong with using this in a lambda, but as you mention, if you do use this (or if you use it implicitly, by calling any nonstatic member function or using a nonstatic member variable) then the garbage collector will keep the object that this refers to alive at least as long as the delegate is alive. Since you pass a lambda to Lazy, this implies that the Repository will be alive at least as long as the Lazy object is alive (even if you never call Lazy.Value).

To demystify it a bit, it helps to look in a disassembler. Consider this code:

class Foo {
    static Action fLambda, gLambda;

    int x;
    void f() {
        int y = 0;
        fLambda = () => ++y;
    }
    void g() {
        int y = 0;
        gLambda = () => y += x;
    }
}

The standard compiler changes this to the following (try to ignore the <> extra angle brackets). As you can see, lambdas that use variables from inside the function body are transformed into classes:

internal class Foo
{
    private static Action fLambda;
    private static Action gLambda;
    private int x;

    private void f()
    {
        Foo.<>c__DisplayClass1 <>c__DisplayClass = new Foo.<>c__DisplayClass1();
        <>c__DisplayClass.y = 0;
        Foo.fLambda = new Action(<>c__DisplayClass.b__0);
    }
    private void g()
    {
        Foo.<>c__DisplayClass4 <>c__DisplayClass = new Foo.<>c__DisplayClass4();
        <>c__DisplayClass.<>4__this = this;
        <>c__DisplayClass.y = 0;
        Foo.gLambda = new Action(<>c__DisplayClass.b__3);
    }

    [CompilerGenerated]
    private sealed class <>c__DisplayClass1
    {
        public int y;
        public void b__0()
        {
            this.y++;
        }
    }
    [CompilerGenerated]
    private sealed class <>c__DisplayClass4
    {
        public int y;
        public Foo <>4__this;
        public void b__3()
        {
            this.y += this.<>4__this.x;
        }
    }

}

If you use this, whether implicitly or explicitly, it becomes a member variable in the compiler-generated class. So the class for f(), DisplayClass1, does not contain a reference to Foo, but the class for g(), DisplayClass2, does.

The compiler handles lambdas in a simpler manner if they don't reference any local variables. So consider some slightly different code:

public class Foo {
    static Action pLambda, qLambda;

    int x;
    void p() {
        int y = 0;
        pLambda = () => Console.WriteLine("Simple lambda!");
    }
    void q() {
        int y = 0;
        qLambda = () => Console.WriteLine(x);
    }
}

This time the lambdas don't reference any local variables, so the compiler translates your lambda functions into ordinary functions. The lambda in p() does not use this so it becomes a static function (called

b__0); the lambda in q() does use this (implicitly) so it becomes a non-static function (called b__2):

public class Foo {
    private static Action pLambda, qLambda;

    private int x;
    private void p()
    {
        Foo.pLambda = new Action(Foo.
b__0);
    }
    private void q()
    {
        Foo.qLambda = new Action(this.b__2);
    }
    [CompilerGenerated] private static void 
b__0()
    {
        Console.WriteLine("Simple lambda!");
    }
    [CompilerGenerated] private void b__2()
    {
        Console.WriteLine(this.x);
    }
    // (I don't know why this is here)
    [CompilerGenerated] private static Action CS$<>9__CachedAnonymousMethodDelegate1;
}

Note: I viewed the compiler output using ILSpy with the option "decompile anonymous methods/lambdas" turned off.