x=0; x=x++; System.out.println(x); why is the output 0?

Answer 1

x=x++ will first assign value of x to x and then increment it.
So again 0 is assigned to x and then incremented value 1 is not stored anywhere.So you get the output as 0 when you print x.
i guess
x=++x;
should work.

Answer 2

you must do:

x=0;
x++;

or

x=0;
x=++x;

(change post-increment to pre-increment)

Answer 3

Of Course! (it's not a loop) he just wanted to change the "0"...

Answer 4

Do you know what this symbol (=) stands for? The left represents the stack memory, the right represents the heap memory. The right should represent the memory address. It is the address of a pointer. X + +, in the heap memory is not out of the stack, you have changed the x pointer, x + + is actually divided into three steps, your wording interrupted the steps of the pointer, if you remove x = these two, you will get you The desired result！

Answer 5

x=0;
x=x++;

in the 2nd line,the variable in the memory will be increment to 1,but as it is a post increment 0 will be fetch at first.,then 0 is assigned to x as assignment is the last operation.,. here the value which previously was 1 will be replaced by 0. hence x=0.

Answer 6

This is post increment, in which original value is first assigned and then it is incremented, unlike in pre increment, value is first incremented and then new value is assigned, same applies for decrement.

Answer 7

I suppose the statement x=x++ assignes to x the value of x before increment (so 0) and also it overrides the ++ increment, since the operations are sequentially:
1) extract the value of x;
2) increment the value of x;
3) assign to x the previously extract value.
Meaning so that x=0. What practically happens is:
tmp = x;
x++;
x = tmp;
I suppose that if the variables were different then the left one would be the previous value of x (0) and also the x variable would increment.

Answer 8

x=0;
x=x++;
here at frst x(right side ) is assign to x(left side) which is 0 and then increase the x(right side) to 1. because it is post increment operation

Answer 9

I also don't get it, isn't

x=0;
x=x++;

and the final value of x will become 1 because of x++ ?