Doubt regarding question from danchisholm on operators

[ani pillai]

I am trying to analyze this question from danchisholm regarding operators:

 
class EBH106 {
  public static void main(String args[]) {
    int a = 1; a += ++a + a++; System.out.print(a);
}}   

I broke down the following piece of code
int a = 1;
a += ++a + a++; System.out.print(a);

to

a = a + (++a + a++);

Now we need to analyse the bracket contents first so I get

a = a+ (2+2); and the value of a at this point is 3

So according to me when I reach the statement

a = a+ (2+2);

I should be remembering the value of a as 3 and the operation should equate to

a = 3 + (4);

But the compiler resolves the value of

a = a+ (2+2);

to

a = 1 + (4);

Why is it so?

[David Marco]

I think the left side of the assignment is guaranteed (and protected from possible changes) until the right side is completely processed, due to the lowest preference of the assignment operators (=, +=, *=, ...)

Greetings.

[Ruben Soto]

The compound assignment operators have a very specific evaluation process:
- Evaluate the value of the variable in the LHS, and save that value.
- Evaluate the RHS expression.
- Perform the operation of the saved value of the LHS with the RHS value.
- Cast the result to the type of the LHS variable.
- Assign the result to the variable in the LHS.

Appart from this, in the RHS, you have ++a + a++. When you have a binary operator, the left operand is evaluated first, and then the right operand is evaluated. Sometimes it can make a difference on the final result (but this time it doesn't.)