Another question from dan's test

[Veena Pointi]

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

In the above code += has lower priority than increment operators.so it should be evaluated the following way
a += 2+2(3)
a=(int)3+2+2
a=7
But the above code prints 5 a
a=(int)1+2+2
a=5
If we exapand += operator before increment operator doesn't that mean += operator is given more priority than ++ operator? ,which ofcourse is not true.Can someone clear my doubt?
Thanks
Veena

[Pradeep bhatt]

From JLS

A compound assignment expression of the form E1 op= E2 is equivalent to E1 = (T)((E1) op (E2)), where T is the type of E1, except that E1 is evaluated only once. Note that the implied cast to type T may be either an identity conversion (�5.1.1) or a narrowing primitive conversion (�5.1.3). For example, the following code is correct:
short x = 3;
x += 4.6;
and results in x having the value 7 because it is equivalent to:

short x = 3;
x = (short)(x + 4.6);

[Veena Pointi]

Pradeep
Actually what i meant is += operator is evaluated first before unary ++ opearator....which is not technically correct,coz ++ opearator has higher priority than +=...
Veena

[Pradeep bhatt]

The += evaluation takes place the last.
a+= b+c
is expanded to
a =a + (b+c)
b+c is evaluated first and then a is added to it => += evaluation takes place last.

[Kashif Memon]

Originally posted by Veena Point:
[QB]

<font size="2">
class U {
  public static void main(String args[]) {
    int a = 1;
    a += ++a + a++;
    System.out.print(a);
  }
}
</font>

In the above code += has lower priority than increment operators.so it should be evaluated the following way
a += 2+2(3)
a=(int)3+2+2
a=7

In the above code += operator is not given higher priority it is evaluated after increment operator.
It is evaluated as under:
a(1) += ++a(2) + (2)a++
a(1) += 4
a = 1 + 4
a=5
I hope it will help you.

[Kashif Memon]

Originally posted by Veena Point:

<font size="2">
class U {
  public static void main(String args[]) {
    int a = 1;
    a += ++a + a++;
    System.out.print(a);
  }
}
</font>

In the above code += has lower priority than increment operators.so it should be evaluated the following way
a += 2+2(3)
a=(int)3+2+2
a=7
Veena

In the above code += operator is not given higher priority it is evaluated after increment operator.
It is evaluated as under:
a(1) += ++a(2) + (2)a++
a(1) += 4
a = 1 + 4
a=5
I hope it will help you.

[Veena Pointi]

Thank you .It helped.I think operands are evaluated first(from left to right) & then expression is evaluated.Am I right?
Veena