Is nameof() evaluated at compile-time?

I wanted to enrich the answer provided by @I3arnon with a proof that it is evaluated at compile-time.

Let's assume i want to print the name of a variable in the Console using the nameof operator:

 var firstname = "Gigi";
 var varname = nameof(firstname);
 Console.WriteLine(varname); // Prints "firstname" to the console

When you check out the MSIL generated you will see that it is equivalent to a string declaration because an object reference to a string gets pushed to the stack using the ldstr operator:

IL_0001: ldstr "Gigi"
IL_0006: stloc.0
IL_0007: ldstr "firstname"
IL_000c: stloc.1
IL_000d: ldloc.1
IL_000e: call void [mscorlib]System.Console::WriteLine(string)

You will notice that declaring the firstname string and using the nameof operator generates the same code in MSIL, which means nameof is as efficient as declaring a string variable.

Yes. nameof() is evaluated at compile-time. Looking at the latest version of the specs:

The nameof expression is a constant. In all cases, nameof(...) is evaluated at compile-time to produce a string. Its argument is not evaluated at runtime, and is considered unreachable code (however it does not emit an "unreachable code" warning).

From nameof operator - v5

You can see that with this TryRoslyn example where this:

public class Foo
{
    public void Bar()
    {
        Console.WriteLine(nameof(Foo));
    }
}

Is compiled and decompiled into this:

public class Foo
{
    public void Bar()
    {
        Console.WriteLine("Foo");
    }
}

Its run-time equivalent is:

public class Foo
{
    public void Bar()
    {
        Console.WriteLine(typeof(Foo).Name);
    }
}

---

As was mentioned in the comments, that means that when you use nameof on type parameters in a generic type, don't expect to get the name of the actual dynamic type used as a type parameter instead of just the type parameter's name. So this:

public class Foo
{
    public void Bar<T>()
    {
        Console.WriteLine(nameof(T));
    }
}

Will become this:

public class Foo
{
    public void Bar<T>()
    {
        Console.WriteLine("T");
    }
}