When passing an object as a parameter to a method, the link must be copied. Then why the output is 0 9 9 , and not 9 9 9 ? It turns out, s1 and s2 refer to different objects?
public class Main { public static void main(String[] args) { Integer s = 0; foo(s, s = 9); System.out.println(s); } static void foo(Integer s1, Integer s2) { System.out.println(s1); System.out.println(s2); } } 
The parameters are calculated from left to right, and therefore s1 and s2 refer to different objects. If you write
foo(s = 9, s); Then get 9 9 9 . Since in this case the operation = executed before the calculation of the second parameter. Your code is equivalent (except for the fact that there are no new variables):
Integer s = 0; // "раскрыл скобки" для foo(s, s = 9); Integer old_s = s; s = 9; foo(old_s, s); // "раскрыл скобки" System.out.println(s); 
= creates a new object, and does not change the value of the old? - user232692= changes the value of the variable to the left and returns the value of the expression to the right. so you can do it like this a = b = c = 123 - Mikhail Vaysmans=9 : 1) allocate memory for the int variable; 2) writing to it 9; 3) we assign THIS address to the reference variable s and it now points to a new memory area; 4) the garbage collector frees the old address? - user232692This happens because you first send to the first argument, and in the second you override the variable by 9. You can override one and the same variable, unlimited count-one time, only if it is not final.
Integer i = 1; i = i * 2; i = i + i; i-=1; 
Source: https://ru.stackoverflow.com/questions/610037/
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