Why does this implicit conversion from int to uint work?

Question

Using Casting null doesn't compile as inspiration, and from Eric Lippert\'s comment:

That demonstrates an interesting case. \"uint x = (int)0;\" w

Answer 1

The following code wil fail with the message "Cannot implicitly convert type 'int' to 'uint'. An explicit conversion exists (are you missing a cast?)"

int y = 0;
uint x = (int)y;

And this will fail with: "Constant value '-1' cannot be converted to a 'uint'"

uint x = (int)-1;

So the only reason uint x = (int)0; works is because the compiler sees that 0 (or any other value > 0) is a compile time constant that can be converted into a uint