问题
The code:
int r=1;
System.out.println(r + (r=2));
The output is: 3. But I expected 4 because I thought the code inside the parentheses is executed first?
回答1:
Official Docs on Operators says
All binary operators except for the assignment operators are evaluated from left to right; assignment operators are evaluated right to left.
So + is evaluated left-to-right,where as assignment operators are evaluated right to left.
回答2:
Use this if you want 4
int r=1;
System.out.println((r=2) + r); // execute + is left-to-right
回答3:
Associativity of + is left-to-right , and the value of the expression (r=2) is 2.
Refer JLS 15.7
The left-hand operand of a binary operator appears to be fully evaluated before any part of the right-hand operand is evaluated.
If the operator is a compound-assignment operator (§15.26.2), then evaluation of the left-hand operand includes both remembering the variable that the left-hand operand denotes and fetching and saving that variable's value for use in the implied binary operation.
If evaluation of the left-hand operand of a binary operator completes abruptly, no part of the right-hand operand appears to have been evaluated.
回答4:
it's like this
(r + (r=2))
(1 + (r=2))
(1 + (2))
(3)
回答5:
The value of statement r=2 is 2. Nested expressions between brackets are processed first though.
For instance:
int r=2;
System.out.println(r * (r=2 + 1));
Output:
6
Why? Because r = 2 + 1 returns 3.
Same as:
int r=2;
System.out.println(r * (2 + 1));
Output still 6 because (2 + 1) is evaluated before multiplication.
来源：https://stackoverflow.com/questions/17808093/java-parentheses-and-assignment