int x = 14; int y = x++; System.out.println(x++); why result is 15?

Question

Postfix: if x = 14 value of x [System.out.println(x++)] will be 14 I got this part but

int x = 14;
int y = x++;
System.out.println(x++);

giving me 15. please, anyone, explain this

Answer 1

Prefix means first increment then assign.
Postfix means first assign then increment.

So y = x++; // first assign so y=14 now x has to be incremented.
When you print it in the next line, it has become 15.

Answer 2

Hi! Your x stay 15 in line:

int y = x++;

it is increased by the x++ Postfix increment by one

Then in line:

System.out.println(x++);

the value 15 is output, which was defined in the previous line and the x was increased by one more time (behind the frame) , so if we add one more line with the command output x. the program will give us the result 16

System.out.println(x);

the result is be 16

https://code.sololearn.com/clNsGINb019D/?ref=app

Answer 3

Because you incremented x when you defined y.

Edit: Try to println x again by it self after the last one.

Answer 4

Yaroslav Vernigora: Result of y would be 14 :)
Thanks, man you made it clear.

Answer 5

//step 1 = x has the value "14".
//step 2 = the value of x remain 14 which is asigned to y.
//step 3 = print the x++ with post increment to value "15"

Answer 6

Question to you: what is result of y?

Answer 7

15