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?
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.
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