++ confusion

[gaurav gupta sitm]

public class MainClass {  
 public static void main(String args[]) {
    int a = 2;
    a += a++; 
    System.out.println(a); 
 }
} 

Output : 4
a value must be increased from 4 to 5 , but why it is not increased ???

[Stephan van Hulst]

Because a++ returns the old value, which is 2, and the a on the left is evaluated before the a++ is performed, so all it does is a = 2 + 2.

[mahesh chandra]

Because a++ returns the old value, which is 2, and the a on the left is evaluated before the a++ is performed, so all it does is a = 2 + 2

@Stephan
I agree that a++ returns old value. statement will be equal to 2+2. What about post increment of variable a. value of variable a should get incremented after evaluation of expression right. Why is it not happening in this case?

[David Byron]

Here's the proposed program:

public class PlusEquals {
    public static void main(String args[]) {
        int a = 2;
        a += a++;
        System.out.println(a);
    }
}

The program shown above behaves very much like this one, which clarifies what's happening:

public class PlusEquals {
    public static void main(String args[]) {
        int a = 2;
        int b = 2;

        a += b++; // After addition to 'a', the expression is further evaluated (++) and the result is placed in variable 'b'

        System.out.println(a + " and " + b);
    }
}

The variable 'a' starts with a value of 2: int a = 2. In the line 'a += a++', the value of 'a' (namely, 2) is assigned additively to the existing value of the variable 'a' (namely, 2), for a new total value of 4. Note that at this point, the assignment of the expression on the right to the variable on the left has already taken place.

Next, the expression on the right is further evaluated (changing from 2 to 3) and then thrown away. Why is it thrown away? Because the assignment of the righthand expression to the lefthand variable has already taken place and there's nowhere else to put the value '3'.

Finally, the current value of the variable 'a' (namely, 4) is printed: 'System.out.println(a)'.

The underlying principle is that there can be only one assignment of value per = operator per expression. Once that condition is satisfied and it's time to increment 'a', there's no place to put the result.

[ankit maini]

public class PlusEquals {
public static void main(String args[]) {
int a = 2;


a = a++; //

System.out.println(a);
}
}

Here the result is 2 not 3.
Remember that a++ or ++a both perform first, difference is that a++ returns the previous value so in this case a++ will increase the value of a by 1 means a++ (a=a+1) is 3 but the it returns the previous value which is 2 so 3 is overwrite by 2 and final result is 2.

[gaurav gupta sitm]

Thanks David .....

[James Elsey]

Hi,

If you are still questioning the incrementing and decrementing operators, I wrote a little guide on them which may be of interest

[anirudh jagithyala]

Step by step process to evaluate the corresponding expression........
int a=2;
a+=a++;

a+=(a++); //value of a is 2
a=a+(a++); //value of a is 2
//remeber the evaluation of expression is from left to right
a=2+(a++); //value of a is 2
a=2+(2); //value of a is 3 as b=a++; then b=2 and a=3
a=2+2;
a=4

The above process is in java as evaluation of expression is left to right



But in some languages like c,c++ the evaluation of expression is from right to leftthen the output would be 5 as follows
a+=a++;
a=a+(a++); //value of a=2
a=a+(2);//value of a=3
a=3+2;
a=5;

in case of assemblers
a+=a++;
a=a+a++;
a=a+a+1;
a=2+2+1;
a=5;


So the output is 4 in case of Java or .Net and the output is 5 with C,C++ -----