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

Question

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

Candace

Always avoid ambiguous statements :) – Arielariela

@Prasoon Saurav Unlike C and C++, Java and C# have strictly defined order of evaluation, so these statements are not ambiguous. – Winterfeed

I know that but still those statements are not(can not be) used for practical purpose so one must avoid it. – Arielariela

https://mcmap.net/q/103396/-java-increment-decrement-operators-how-they-behave-what-39-s-the-functionality – Oech

@PeteKirkham It's over six years later, but I still want to point out that "ambiguous", in this situation, is ambiguous -- it could mean "the compiler doesn't know what to put", or it could mean "The programmer has no idea what it means". – Hendry

@QPaysTaxes Java programmers should know what pre and post-increment mean and that statements are evaluated left to right, so if the programmer really has no idea what it means then they aren't a Java programmer. Which isn't to say it's a clear way of expressing the semantics, but it not in any way ambiguous. – Winterfeed

i=++a + ++a + a++; => i=7 + 8 + 5; (a=8) since post increment has highest precedence, which means a++ should executed first. – Tampa

Understand that Ankit is trying to contrast the two ways of increment in his example code; just in case someone is wondering like me: putting the example code exactly as written in Eclipse gives 38 and 29. – Orling

Answer 1

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.

Subjection

Are you sure a == 9 in the second one? – Winterfeed

i=++a + ++a + a++; => i=7 + 8 + 5; (a=8) since post increment has highest precedence, does a++ is executed first? – Tampa

complicated example to something that is easy to explain. – Solace

Is this answer is the same for c# and c++? – Dielectric

@Dennis b = a++ results in b = 1, a = 2, then c = ++b results in c = 2, b = 2 – Sommelier

Answer 2

++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.

Ardene

Instead of "uses", I think a less confusing word would have been "evaluates". – Sallyanne

Answer 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;

Blevins

Answer 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.

Crowberry

these statements work right to left or left to right? – Reed

Answer 5

++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.

Swats

@KlasLindbäck commutative means that you can swap both expressions and still get the same result. So a++ + ++a == ++a + a++ (5 + 7 == 6 + 6; a == 7 at the end). – Swats

Answer 6

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

Hispanicism

a is 39 in c++. Why so? – Penalize

Answer 7

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

Hedgehog

yeah, this is the only right answer corresponding to the crappy code provided. Other ppl have interpreted the code to make it a legitimate valid question. – On

Answer 8

++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!

Dipietro

Answer 9

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.

Millpond

Answer 10

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

Billmyre

Answer 11

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.

Stature

Answer 12

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

Everywhere

Answer 13

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)

Lysimachus

Answer 14

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.

Coopt