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");
}
}
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