How do the post increment (i++) and pre increment (++i) operators work in Java?

Solution 1:

++a increments and then uses the variable.
a++ uses and then increments the variable.

If you have

a = 1;

and you do

System.out.println(a++); //You will see 1

//Now a is 2

System.out.println(++a); //You will see 3

codaddict explains your particular snippet.

Solution 2:

Does this help?

a = 5;
i=++a + ++a + a++; =>
i=6 + 7 + 7; (a=8)

a = 5;
i=a++ + ++a + ++a; =>
i=5 + 7 + 8; (a=8)

The main point is that ++a increments the value and immediately returns it.

a++ also increments the value (in the background) but returns unchanged value of the variable - what looks like it is executed later.

Solution 3:

In both cases it first calculates value, but in post-increment it holds old value and after calculating returns it

++a

1. a = a + 1;
2. return a;

a++

1. temp = a;
2. a = a + 1;
3. return temp;

Solution 4:

i = ++a + ++a + a++;

is

i = 6 + 7 + 7

Working: increment a to 6 (current value 6) + increment a to 7 (current value 7). Sum is 13 now add it to current value of a (=7) and then increment a to 8. Sum is 20 and value of a after the assignment completes is 8.

i = a++ + ++a + ++a;

is

i = 5 + 7 + 8

Working: At the start value of a is 5. Use it in the addition and then increment it to 6 (current value 6). Increment a from current value 6 to 7 to get other operand of +. Sum is 12 and current value of a is 7. Next increment a from 7 to 8 (current value = 8) and add it to previous sum 12 to get 20.