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c#nullablevalue-typeboxingnullable-reference-types
franz1
franz1

Reputation: 351

Making value type variables nullable by using "?": does it imply boxing?

Given the following assumptions in C#:

  1. boxing and unboxing let me convert any value type variable to an object type, which is a reference type (therefore it's also nullable), like in the example:
int i = 123;
object box = i;
  1. The "?" operator let me convert a not-nullable integer into a nullable variable, like in the example:

int? n = 0;

My Question: for only reference type variables are nullable, can I say that in the second example I'm doing implicit boxing? In other terms, when I use the "?" operator to make nullable an integer, is there boxing operation implied as well (even if it's not explicit)?

Upvotes: 0

Views: 209

Answers (1)

JonasH
JonasH

Reputation: 36371

int? is suggar for Nullable<T> See documentation. If we look at the signature of this we see:

public struct Nullable<T> where T : struct
{
...
    public override bool Equals(object other)
    {
      if (!this.hasValue)
        return other == null;
      return other != null && this.value.Equals(other);
    }

Since it is a struct the value will not be boxed.

If you need to compare values, n.Equals(1) would cause boxing of the argument. I cannot find any documentation about the equality operator ==, but I think it would be fairly safe to assume that it should not cause boxing.

Upvotes: 1

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