Java: Prefix/postfix of increment/decrement operators

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" 

Think of ++i and i++ as similar to i = i+1. But it is not the same. The difference is when i gets the new increment.

In ++i , the increment happens immediately.

But if i++ is there, the increment will happen when the program goes to the next line.

Look at the code here.

int i = 0; while(i < 10){ System.out.println(i); i = increment(i); } private int increment(i){ return i++; } 

This will result in a nonending loop. Because i will be returned with the original value and after the semicolon, i will get incremented, but the returned value has not been. Therefore i will never actually returned as an incremented value.

Why wouldn't the variable have been updated?

• Postfix: passes the current value of i to the function and then increments it.
• Prefix: increments the current value and then passes it to the function.

The lines where you don't do anything with i make no difference.

Notice that this is also true for assignments:

i = 0; test = ++i; // 1 test2 = i++; // 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).

An example of how the actual operators are implemented:

class Integer { private int __i; function Integer ++() { // Prefix operator i.e. ++x __i += 1; // Increment return this; // Return the object with the incremented value } function Integer ++(Integer x) { // Postfix operator, i.e., x++ __i+=1; // Increment return x; // Return the original object } } 

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); } } 

This may be 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. } } 

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

This might help... it also took my to understand this puzzle.

public class Main { public static void main(String[] args) { int x = 5; int y = 5; System.out.println(++x); System.out.println(y++); System.out.println("------"); System.out.println(x); System.out.println(y); } } 

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