This null-check will always throw an exception because the expression will always evaluate to 'null'DRT-W1554

The analyzer produces this diagnostic when the null check operator (!) is used on an expression whose value can only be null. In such a case the operator always throws an exception, which likely isn't the intended behavior.

Example

The following code produces this diagnostic because the function g will always return null, which means that the null check in f will always throw:

void f() {
  g()!;
}

Null g() => null;

Common fixes

If you intend to always throw an exception, then replace the null check with an explicit throw expression to make the intent more clear:

void f() {
  g();
  throw TypeError();
}

Null g() => null;