Semantics of pre- and postfix "++" operator in Java [duplicate]

Question

This question already has answers here:

Java: Prefix/postfix of increment/decrement operators? (11 answers)

Closed 8 years ago.

I wondering to know why this snippet of code give output 112
How this last digit 2 was creating?

public static void main(String[] args) {
    int i = 0;
    System.out.print(++i);
    System.out.print(i++);
    System.out.print(i);

Why does this happen?

Answer 1

Your snippet it's translated as

int i = 0;
i = i + 1; // 1
System.out.print(i); // 1
System.out.print(i); // 1
i = i + 1; // 2
System.out.print(i); // 2

That's why the final result it's a 2.

++i it's incrementing the variable before being called by the print method and i++ it's incrementing the variable after the method execution.

Answer 2

i++ is the post-increment operator, which has expression value the old value of i, but a side-effect of incrementing i. The value is 1, but it leaves i changed to 2.

Answer 3

When we use post or pre increment operator it increases the value.

Post increment operator (i++) assigns the value first, then increments it. Pre increment operator (++i) increments first then assigns the value. They both behave like this :

int i=0;
i=i++;
System.out.println(i); //1
i=++i;
System.ou.println(i); //1

Answer 4

When this code runs:

public static void main(String[] args) {
    int i = 0;                //i=0;
    System.out.print(++i);    // increments i to i=1, and prints i
    System.out.print(i++);    // prints i and then increments it to i=2
    System.out.print(i);      // prints i, i.e. 2
}

Answer 5

i is initially 0, then it is pre-incremented and printed so you have the first 1, then it is printed again and you have the second 1, then post-incremented, then printed for the last time and you have the 2

Answer 6

Simply;

In post increment, the increment is done after the variable is read.

In pre increment, the variable value is incremented first, then used in the expression.

Answer 7

You are applying two increments on i. The initial value was 0 so after two increments (++i and i++ )it will become 2.

Both i++ and ++i are incrementing the value of i by one.

They are similar to

i = i+1;

but the ++i one increments the value of i then uses it, so 0 becomes 1 and printed out, while the i++ first uses the value and then increments the value of i, so the printed value is 1 and then it becomes 2 hence the last digit(the final value of i) is 2.