what will be the reason behind the output of java code?

Question

Consider the following simple java code

public static void main(string args[])
{
  int x=10;
  int y;

  y=x++;
  y=y--;

  System.out.println(y);
}

here output is 10. but actually y variable decrements. but according to me output should be 9. what is the reason?

`y=y--` is the same as `y=y`. It has no effect on the value of `y` **at the end** of the assignment. — barak manos, Aug 23 '14 at 10:14
`y = x++;` the same `y = x; x += 1;` , `y = y--; (y = y , y -=1)` — , Aug 23 '14 at 10:15

score 3 · Answer 1 · answered Aug 23 '14 at 10:17

The postfix-increment operator works as follows:

The postfix-decrement operator works as follows:

Therefore, the following piece of code works as follows:

x = 10;  // x == 10
y = x++; // x == 11, y == 10
y = y--; // y ==  9, y == 10

As you can see, y = y-- is equivalent to y = y.

It has no effect on the value of y at the end of the operation.

score 1 · Answer 2 · answered Aug 23 '14 at 10:18

1

You need to understand about prefix and postfix operators.

y=x++ means assign x to y and then increment x by 1;

y=++x means increment x and then assign the incremented value to y.

If you understand this difference then its obvious what the code does.

answered Aug 23 '14 at 10:18

Nazgul

I don't know how this answer is different from above answer so it got 2 votes up! +1 – Aug 23 '14 at 10:22
I don't think what you stated is equivalent to barak's answer. Your answer implies that the decrement occurs after the assignment. According to your statement, `y=y--` is equivalent to `y=y; y=y-1;` which is not the same as what the other answers are saying... – Tung Aug 23 '14 at 10:30
explaining something to a beginner in sample code is often confusing as against explaining the same thing in plain English. – Nazgul Aug 23 '14 at 12:12

score 0 · Answer 3 · answered Aug 23 '14 at 10:21

The post-decrement operator returns the old value of the variable. So when you write

y = y--;

it decrements y, then assigns the old value back to y. It's as if you wrote:

old_y = y; // Save old value
y = y - 1; // perform y-- decrement
y = old_y; // Assign the saved value

score 0 · Answer 4 · answered Aug 23 '14 at 10:26

Here's whats happening

int x=10;

declare a vaiable x and assign 10 to it.

int y;

declare a variable y

y=x++;

increment x therefore x=11 and return its old value therefore y=10

y=y--;

decrement y therefore y=9 at current point, and return its old value which is 10 and is caught by y therefore y=10 now.

System.out.println(y);

Output

score 0 · Answer 5 · answered Aug 23 '14 at 11:41

x = 10; // x == 10

y = x++; //Here First Y is asigned with value 10 and increments x by 1

y = y--;//Here First Y is asigned with value 10 and decrements y by 1

Because If we use Postincrement or postdecrement First it asigns value to Variable then it performs increment or decrement .

5 Answers5