Java: Prefix/postfix of increment/decrement operat

Question

From the program below or here, why does the last call to System.out.println(i) print the value 7?

class PrePostDemo {
     public static void main(String[] args){
          int i = 3;
          i++;
          System.out.println(i);    // "4"
          ++i;             
          System.out.println(i);    // "5"
          System.out.println(++i);  // "6"
          System.out.println(i++);  // "6"
          System.out.println(i);    // "7"
     }
}

Answer 1

System.out.println(i++);  // "6"

This sends println the value I had prior to this line of code (6), and then increments I (to 7).

Answer 2

This is my answer. Some of you may find it easy to understand.

package package02;

public class C11PostfixAndPrefix {

    public static void main(String[] args) {
        // In this program, we will use the value of x for understanding prefix 
        // and the value of y for understaning postfix. 
        // Let's see how it works. 

        int x = 5; 
        int y = 5; 

        Line 13:   System.out.println(++x);  // 6   This is prefixing. 1 is added before x is used. 
        Line 14:   System.out.println(y++);  // 5   This is postfixing. y is used first and 1 is added. 

        System.out.println("---------- just for differentiating");

        System.out.println(x);  // 6   In prefixing, the value is same as before {See line 13}
        System.out.println(y);  // 6   In postfixing, the value increases by 1  {See line 14} 

        // Conclusion: In prefixing (++x), the value of x gets increased first and the used 
        // in an operation. While, in postfixing (y++), the value is used first and changed by
        // adding the number. 
    }
}

Answer 3

I know this has been answered, but thought another explanation may be helpful.

Another way to illustrate it is:

++i will give the result of the new i, i++ will give the result of the original i and store the new i for the next action.

A way to think of it is, doing something else within the expression. When you are printing the current value of i, it will depend upon whether i has been changed within the expression or after the expression.

    int i = 1;
result i = ++i * 2 // result = 4, i = 2

i is evaluated (changed) before the result is calculated. Printing i for this expression, shows the changed value of i used for this expression.

result i = i++ * 2 // result = 2, i = 2

i is evaluated after the result in calculated. So printing i from this expression gives the original value of i used in this expression, but i is still changed for any further uses. So printing the value for i immediately after the expression, will show the new incremented value of i. As the value of i has changed, whether it is printed or used.

result i = i++ * 2 // result = 2, i = 2
System.out.println(i); // 2

---

If you kept a consistent pattern and included print lines for all the values:

  int i = 3; 
System.out.println(i);    //  3
System.out.println(i++);  //  3
System.out.println(i);    // "4"
System.out.println(++i);  //  5          
System.out.println(i);    // "5"
System.out.println(++i);  // "6"
System.out.println(i++);  // "6"
System.out.println(i);    // "7"

Answer 4

It prints 7 for the last statement, cos in the statement above, it's value is 6 and it's incremented to 7 when the last statement gets printed

Answer 5

Maybe you can understand better Prefix/postfix with this example.

public class TestPrefixPostFix 
{
    public static void main (String[] args)
    { 
        int x=10;
        System.out.println( (x++ % 2 == 0)?"yes "+ x: " no "+x);
        x=10;
        System.out.println( (++x % 2 == 0)?"yes "+ x: " no "+x);
    }
}    

Answer 6

Well think of it in terms of temporary variables.

i =3 ;
i ++ ; // is equivalent to:   temp = i++; and so , temp = 3 and then "i" will increment and become     i = 4;
System.out.println(i); // will print 4

Now,

i=3;
System.out.println(i++);

is equivalent to

temp = i++;  // temp will assume value of current "i", after which "i" will increment and become i= 4
System.out.println(temp); //we're printing temp and not "i"