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125

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3

This MSDN reference seems to indicate that when an int? (or any Nullable<T>) has a value, it's always boxed (and hence a much less efficient store of data, memory-wise than an int). Is that the case?

+10  A: 

That page refers to when you are boxing the Nullable<T> struct, not the values inside the struct itself.

There is no boxing involved in storing a nullable type until you try boxing the nullable itself:

int? a = 42; // no boxing
int? n = null; // no boxing

object nObj = n; // no boxing
object aObj = a; // only now will boxing occur

This behavior is no different than when boxing a regular value type (with the exception of handling the null case) so it is really to be expected.

Justin Niessner
+2  A: 

That's not the case. Nullable<T> is generic, so it holds the real int or bool.

The MSDN page is talking about what happens when you box a Nullable<int> or Nullable<bool>. If you never assign your nullable struct to an object variable, you won't incur boxing overhead.

Tim Robinson
+2  A: 

No. The Nullable object is a generic struct, and internally handles the value of T without boxing.

BC