Does passing a value type in an “out” parameter cause the variable to be boxed?

Question

I\'m aware that boxing and unboxing are relatively expensive in terms of performance. What I\'m wondering is:

Does passing a value type to a method\'s out

Answer 1

No, there is no Boxing (required/involved).

When you do Box a variable, changes to the boxed instance do not affect the original. But that is exactly what out is supposed to do.

The compiler 'somehow' constructs a reference to the original variable.