Java priority of multiple operators

This is more a theoretical problem to understand Java's evaluation of arithmetic operations Since and – have the same priority, I don't quite understand how Java evaluates the following expression (there are multiple operators between two operands)

public static void main(String[] args) {
    int a = 1;
    int b = 2;
    System.out.println(a+-b);    // results in -1
    System.out.println(a-+b);    // results in -1
    System.out.println(a+-+b);   // results in -1
    System.out.println(a-+-b);   // results in  3
    System.out.println(a-+-+b);  // results in  3
    System.out.println(a+-+-b);  // results in  3
    System.out.println(a-+-+-b); // results in -1
    System.out.println(a+-+-+b); // results in  3
}

From the Java 8 Language Specification (§ 15.8.2):

I also noticed that every time #perators are even, the result is the same, and the order is not important However, when #operators are odd, this does not necessarily affect the results For example There is also one of the following two expressions – but the result is different

System.out.println(a-+-b);   // results in 3
System.out.println(a-+-+-b); // results in -1

With all this information, I still can't see this model or how it works

Solution

How do you evaluate this mathematically?

a - + - b

Adding some parentheses helps:

a - (+ (-b))

We can do this because it does not violate the priority rule

Then we can start to reduce: (- b) is really - B, and - B is really b, so the result is 1, 2 = 3

Let's look at the second:

a - + - + - b
a - (+ (- (+ (-b))))
a - (+ (- (-b)))
a - (+ b)
a - b
1 - 2 = -1

Such a simple mathematical rule is natural