i and i=i++ in for loop java

Question

What is the logic behind this behaviour?

 int i=0;
    for(int k=0;k<10;k++){
    i++;
    }
    System.out.println("i="+i);

Output=10; //Exepcted



 int i=0;
    for(int k=0;k<10;k++){
    i=i++;
    }
    System.out.println("i="+i);

Output=0; //Surprised :)

Can anybody throw some light on above functionality?

http://stackoverflow.com/questions/7911776/what-is-x-after-x-x — AllTooSir, May 02 '13 at 10:21

score 14 · Accepted Answer · edited May 23 '17 at 12:02

14

See this brilliant answer:

x = x++;

is equivalent to

int tmp = x;
x++;
x = tmp;

From this question.

edited May 23 '17 at 12:02

Community

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answered May 02 '13 at 10:23

Maroun

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3

anybody trying to understand `x=x++` keep in mind that post-increment has given priority over the assignment and x++ increments but returns old value. – chirag soni Jan 28 '20 at 16:27

score 7 · Answer 2 · answered May 02 '13 at 10:21

i = i++ is a postfix increment operator - it increments i, then returns it to its original value (because the i++ essentially "returns" the value of i before it was incremented.)

i = ++i would work since it's a prefix increment operator, and would return the value of I after the increment. However, you probably just want to do i++ there without any extra assignment as you do in the first run - it's (essentially) shorthand as it is for i = i+1.

score 5 · Answer 3 · answered May 02 '13 at 10:22

What is happening happens because java is pass-by-value.

In the first loop, i is getting incremented in the i++ statement, however, in the second loop what is happening is that i gets pointed to a new memory location that stores the value of i (in this case 0) and then increments the old location.

To visualise:

i => 0x00000001 // 0

for() {
    i => 0x00000002 <- 0  // store old i value (0) in new location
    0x00000001++          // Increment the value stored at the old location

    // Cause there is no longer a reference to 0x00000001, 
    // it will get garbage collected and you will be left with
    // i => 0x00000002

And it will keep doing that, assigning the old value to a new location and incrementing the old value for each pass of the loop

score 2 · Answer 4 · answered May 02 '13 at 10:22

2

i=i++; will never increment i because the ++ is processed after the i=i.

you could see it like this:

int i=0;
for(int k=0;k<10;k++){
    int j = 0;
    i = j;
    j = j + 1;
}

answered May 02 '13 at 10:22

Philipp Sander

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score 2 · Answer 5 · answered May 02 '13 at 10:23

2

i = i++;

is equivalent to,

int temp = i; // temp = 0
i++; // i=1
i = temp; // i = 0

answered May 02 '13 at 10:23

AllTooSir

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score 1 · Answer 6 · answered May 02 '13 at 10:22

1

The ++ operator is processed after the assigment,

if you changed it i=++i; you'd probably get the behaviour you expected

answered May 02 '13 at 10:22

paul

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Well yes. But you shouldn't recommend a bad solution. Better solutions are `i++;` or `i +=1;` or `i = i + 1;`. – Stephen C May 02 '13 at 11:15
yes agreed, `i=++i;` would never be used in practice – paul May 02 '13 at 11:18

score 1 · Answer 7 · answered May 02 '13 at 10:23

1

In the first option , you are incrementing the i by using i++ , which is equivalent to i=i+1, so it increases the value of i to 10.
but in the second option , you are assining i a new value , hence getting the same value everytime.

answered May 02 '13 at 10:23

Vineet Singla

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MemLeak · Answer 8 · 2013-05-02T13:32:20.777

1

i=i++;

returns i and increments. so the increment is lost... look at this pseudo code

x = i++ will break the operation in following steps

x = i;
i++;

in your case , it returns 0 increments to 1 (but the increment is lost)

edited May 02 '13 at 13:32

answered May 02 '13 at 10:24

MemLeak

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i and i=i++ in for loop java

8 Answers8

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