Why does this Java code compile?

Question

In method or class scope, the line below compiles (with warning):

int x = x = 1;

In class scope, where variables get their default valu

Answer 1

The line of code does not compile with a warning because of how the code actually works. When you run the code int x = x = 1, Java first creates the variable x, as defined. Then it runs the assignment code (x = 1). Since x is already defined, the system has no errors setting x to 1. This returns the value 1, because that is now the value of x. Therefor, x is now finally set as 1.
Java basically executes the code as if it was this:

int x;
x = (x = 1); // (x = 1) returns 1 so there is no error

However, in your second piece of code, int x = x + 1, the + 1 statement requires x to be defined, which by then it is not. Since assignment statements always mean the code to the right of the = is run first, the code will fail because x is undefined. Java would run the code like this:

int x;
x = x + 1; // this line causes the error because `x` is undefined