In Java, which gets executed first, "+" or "++"?

Question

I tried the following code in Java

t1 = 5; t2 = t1 + (++t1); System.out.println (t2); 

My view is since ++ has a higher precedence than +, the above becomes

t2 = t1 + (++t1); t2 = t1 + 6;      // t1 becomes 6 here t2 = 6 + 6; t2 = 12; 

However, I get the answer 11 for t2. Can someone explain?

Answer 1

You are nearly correct but you are subtly misunderstanding how the precedence rules work.

Compare these two cases:

int t1 = 5; int t2 = t1 + (++t1); System.out.println (t2);  t1 = 5; t2 = (++t1) + t1; System.out.println (t2); 

The result is:

11 12 

The precedence does indeed say to evaluate the ++ before the +, but that doesn't apply until it reaches that part of the expression.

Your expression is of the form X + Y Where X is t1 and Y is (++t1)

The left branch, i.e. X, is evaluated first. Afterwards the right branch, i.e. Y, is evaluated. Only when it comes to evaluate Y the ++ operation is performed.

The precedence rules only say that the ++ is "inside" the Y expression, they don't say anything about the order of operations.