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

Can you explain to me the output of this Java code?

int a=5,i;

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

System.out.println(a);
System.out.println(i);

The output is 20 in both cases

pre-increment and post increment are equivalent if not in an expression

int j =0;
int r=0         
for(int v = 0; v<10; ++v) { 
          ++r;
          j++;
          System.out.println(j+" "+r);
  }  
 1 1  
 2 2  
 3 3       
 4 4
 5 5
 6 6
 7 7
 8 8
 9 9
10 10

I believe however if you combine all of your statements and run it in Java 8.1 you will get a different answer, at least that's what my experience says.

The code will work like this:

int a=5,i;

i=++a + ++a + a++;            /*a = 5;
                                i=++a + ++a + a++; =>
                                i=6 + 7 + 7; (a=8); i=20;*/

i=a++ + ++a + ++a;           /*a = 5;
                                i=a++ + ++a + ++a; =>
                                i=8 + 10 + 11; (a=11); i=29;*/

a=++a + ++a + a++;            /*a=5;
                                a=++a + ++a + a++; =>
                                a=12 + 13 + 13;  a=38;*/

System.out.println(a);        //output: 38
System.out.println(i);         //output: 29

when a is 5, then a++ gives a 5 to the expression and increments a afterwards, while ++a increments a before passing the number to the expression (which gives a 6 to the expression in this case).

So you calculate

i = 6 + 7 + 7
i = 5 + 7 + 8

++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.

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.

In the above example

int a = 5,i;

i=++a + ++a + a++;        //Ans: i = 6 + 7 + 7 = 20 then a = 8 

i=a++ + ++a + ++a;        //Ans: i = 8 + 10 + 11 = 29 then a = 11

a=++a + ++a + a++;        //Ans: a = 12 + 13 + 13 = 38

System.out.println(a);    //Ans: a = 38

System.out.println(i);    //Ans: i = 29

Presuming that you meant

int a=5; int i;

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

System.out.println(i);

a=5;

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

System.out.println(i);

a=5;

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

System.out.println(a);

This evaluates to:

i = (6, a is now 6) + (7, a is now 7) + (7, a is now 8)

so i is 6 + 7 + 7 = 20 and so 20 is printed.

i = (5, a is now 6) + (7, a is now 7) + (8, a is now 8)

so i is 5 + 7 + 8 = 20 and so 20 is printed again.

a = (6, a is now 6) + (7, a is now 7) + (7, a is now 8)

and after all of the right hand side is evaluated (including setting a to 8) THEN a is set to 6 + 7 + 7 = 20 and so 20 is printed a final time.

++a increments a before it is evaluated. a++ evaluates a and then increments it.

Related to your expression given:

i = ((++a) + (++a) + (a++)) == ((6) + (7) + (7)); // a is 8 at the end
i = ((a++) + (++a) + (++a)) == ((5) + (7) + (8)); // a is 8 at the end

The parenteses I used above are implicitly used by Java. If you look at the terms this way you can easily see, that they are both the same as they are commutative.

I believe you are executing all these statements differently
executing together will result => 38 ,29

int a=5,i;
i=++a + ++a + a++;
//this means i= 6+7+7=20 and when this result is stored in i,
//then last *a* will be incremented <br>
i=a++ + ++a + ++a;
//this means i= 5+7+8=20 (this could be complicated, 
//but its working like this),<br>
a=++a + ++a + a++;
//as a is 6+7+7=20 (this is incremented like this)

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;

Pre-increment means that the variable is incremented BEFORE it's evaluated in the expression. Post-increment means that the variable is incremented AFTER it has been evaluated for use in the expression.

Therefore, look carefully and you'll see that all three assignments are arithmetically equivalent.

++a is prefix increment operator:

• the result is calculated and stored first,
• then the variable is used.

a++ is postfix increment operator:

• the variable is used first,
• then the result is calculated and stored.

Once you remember the rules, EZ for ya to calculate everything!

a=5; i=++a + ++a + a++;

is

i = 7 + 6 + 7

Working: pre/post increment has "right to left" Associativity , and pre has precedence over post , so first of all pre increment will be solve as (++a + ++a) => 7 + 6 . then a=7 is provided to post increment => 7 + 6 + 7 =20 and a =8.

a=5; i=a++ + ++a + ++a;

is

i=7 + 7 + 6

Working: pre/post increment has "right to left" Associativity , and pre has precedence over post , so first of all pre increment will be solve as (++a + ++a) => 7 + 6.then a=7 is provided to post increment => 7 + 7 + 6 =20 and a =8.