What does exclamation mark mean before invoking a method in C# 8.0? [duplicate]

Answer 1

It means non-nullable.

For example:

int can not be null by its nature but we can declare nullable int like:

Nullable<int> x;

or

int? x;

However, some classes can be null even if we don't say so.

for example

string x = null;

and we don't need to declare it as:

string? x;

To make sure that declared variable cannot be null (non-nullable), ! is used, so:

string! x = "somevalue";

it means that x can not accept null as a value.

https://docs.microsoft.com/en-us/archive/msdn-magazine/2018/february/essential-net-csharp-8-0-and-nullable-reference-types

Answer 2

This is called the null-forgiving operator and is available in C# 8.0 and later. It has no effect at run time, only at compile time. It's purpose is to inform the compiler that some expression of a nullable type isn't null to avoid possible warnings about null references.

In this case it tells the compiler that Entity isn't null.

Answer 3

! is the Null-Forgiving Operator. To be specific it has two main effects:

• it changes the type of the expression (in this case it modifies Entity) from a nullable type into a non-nullable type; (for example, object? becomes object)

• it supresses nullability related warnings, which can hide other conversions

This seems to come up particularly with type parameters:

IEnumerable<object?>? maybeListOfMaybeItems = new object[] { 1, 2, 3 };

// inferred as IEnumerable<object?>
var listOfMaybeItems = maybeListOfMaybeItems!;

// no warning given, because ! supresses nullability warnings
IEnumerable<object> listOfItems = maybeListOfMaybeItems!;

// warning about the generic type change, since this line does not have !
IEnumerable<object> listOfItems2 = listOfMaybeItems;

Answer 4

This would be the null forgiving operator.
It tells the compiler "this isn't null, trust me", so it does not issue a warning for a possible null reference.

In this particular case it tells the compiler that Entity isn't null.