Java increment and assignment operator
increment, java, post-increment, pre-increment
Solution
No, the printout of 10 is correct. The key to understanding the reason behind the result is the difference between pre-increment `++x` and post-increment `x++` compound assignments. When you use pre-increment, the value of the expression is taken after performing the increment. When you use post-increment, though, the value of the expression is taken before incrementing, and stored for later use, after the result of incrementing is written back into the variable.
Here is the sequence of events that leads to what you see:
- `x` is assigned `10`
- Because of `++` in post-increment position, the current value of `x` (i.e. `10`) is stored for later use
- New value of `11` is stored into `x`
- The temporary value of `10` is stored back into `x`, writing right over `11` that has been stored there.
Problem
I am confused about the post ++ and pre ++ operator , for example in the following code ``` int x = 10; x = x++; sysout(x); ``` will print 10 ? It prints 10,but I expected it should print 11 but when I do ``` x = ++x; instead of x = x++; ``` it will print eleven as I expected , so why does x = x++; doesn't change the the value of x ?