I don`t understand how Java is progressing this arithmetic expression
int x = 1;
int y = 1;
x += y += x += y;
System.out.println("x=" + x + " y=" + y);
With Java I get x = 4 and y = 3. But in C, Perl, Php I get x=5 and y = 3 On the paper I also get x = 5 and y = 3
This is the nasty part, obviously:
x += y += x += y;
This is executed as:
int originalX = x; // Used later
x = x + y; // Right-most x += y
y = y + x; // Result of "x += y" is the value stored in x
x = originalX + y; // Result of "y += x" is the value stored in y
So:
x y
(Start) 1 1
x = x + y 2 1
y = y + x 2 3
x = originalX + y 4 3
The important part is the use of originalX here. The compound assignment is treated as:
x = x + y
and the first operand of + is evaluated before the second operand... which is why it takes the original value of x, not the "latest" one.
From JLS section 15.16.2:
If the left-hand operand expression is not an array access expression, then:
First, the left-hand operand is evaluated to produce a variable. If this evaluation completes abruptly, then the assignment expression completes abruptly for the same reason; the right-hand operand is not evaluated and no assignment occurs.
Otherwise, the value of the left-hand operand is saved and then the right-hand operand is evaluated. If this evaluation completes abruptly, then the assignment expression completes abruptly for the same reason and no assignment occurs.
When in doubt, consult the language specification - and never assume that just because two languages behave differently, one of them is "wrong". So long as the actual behaviour matches the specified behaviour for each language, all is well - but it does mean you need to understand the behaviour of each language you work with, of course.
That said, you should clearly avoid horrible code like this in the first place.
it starts from right to left
1. x += y gives x = 2, y = 1
2. y += ( x +=y ) gives x = 2, y = 3
3. x += ( y += x += y ) gives x = 4, y = 3
so, it's all consistent
x go back to being 1? It should show x = 2, y = 3. (That's what you'd get if the whole statement was just y += x += y;) At that point you need to explain why the final value is 4... which takes a bit more work.x += ... gives an answer of 4, not 5. Naively, one might think that if the value of x is 2, and the value of y is 3, then x += y will give x = 5. The important part is that x is evaluated before any of the rest happens and that's the value which is added to in the last step.