What is changed by applying nullable Operator on value type datatype that now it can store null.
Technically a null value is a reference (called a pointer in some languages) to an empty area of memory. Reference variables (variables that contain an address for data rather than the data itself) are always nullable , which means they are automatically capable of having a null value.
A null coalescing operator, in C#, is an operator that is used to check whether the value of a variable is null. It is represented by the symbol "??". The null coalescing operator allows for the selection of the first non-null value from a pair of values.
Nullable types work as a connector between a database and C# code to provide a way to transform the nulls to be used in C# code. Null Coalescing operators simplify the way to check for nulls and shorten the C# code when the developer is dealing with nulls.
You typically use a nullable value type when you need to represent the undefined value of an underlying value type. For example, a Boolean, or bool , variable can only be either true or false . However, in some applications a variable value can be undefined or missing.
As others have said, "?" is just shorthand for changing it to Nullable<T>
. This is just another value type with a Boolean flag to say whether or not there's really a useful value, or whether it's the null value for the type. In other words, Nullable<T>
looks a bit like this:
public struct Nullable<T>
{
private readonly bool hasValue;
public bool HasValue { get { return hasValue; } }
private readonly T value;
public T value
{
get
{
if (!hasValue)
{
throw new InvalidOperationException();
}
return value;
}
}
public Nullable(T value)
{
this.value = value;
this.hasValue = true;
}
// Calling new Nullable<int>() or whatever will use the
// implicit initialization which leaves value as default(T)
// and hasValue as false.
}
Obviously in the real code there are more methods (like GetValueOrDefault()
) and conversion operators etc. The C# compiler adds lifted operators which effectively proxy to the original operators for T
.
At the risk of sounding like a broken record, this is still a value type. It doesn't involve boxing... and when you write:
int? x = null;
that's not a null reference - it's the null value of Nullable<int>
, i.e. the one where hasValue
is false
.
When a nullable type is boxed, the CLR has a feature whereby the value either gets boxed to a null reference, or a plain boxed T. So if you have code like this:
int? x = 5;
int y = 5;
object o1 = x;
object o2 = y;
The boxed values referred to by o1
and o2
are indistinguishable. You can't tell that one is the result of boxing a nullable type.
The ?
syntax is syntactic sugar to the Nullable<T>
struct.
In essence, when you write int? myNullableInt
, the compiler changes it to Nullable<int> myNullableInt
.
From MSDN (scroll down to "Nullable Types Overview"):
The syntax T? is shorthand for Nullable, where T is a value type. The two forms are interchangeable.
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