why is this the output of the following loop?

Question

why am I getting the following output: 4 8 12 16 20

int i, j = 1, k;
  for (i = 0; i < 5; i++)
  {
     k = j++ + ++j;
     Console.Write(k + " ");
  }

Do read about increment operator and its effect on the expression it is on depending on its position. — Pablo Santa Cruz, Mar 21 '12 at 00:42
Is this a real question? I mean if you have used the pre and post increment operator then you must have some idea of what they do right? — EdChum, Mar 21 '12 at 00:47

score 2 · Answer 1 · edited May 23 '17 at 12:14

2

Well k = j++ (+) ++j

j++ will increment the value of j, but return the pre-incremented value.
++j will increment the value of j, and then return the incremented value.

j++ = 2 but really returns (1)

although, as soon as you hit ++j, you are incrimenting the real value of j which currently is 2.

++j = 3

1 + 3 = 4

What is the difference between ++i and i++?

edited May 23 '17 at 12:14

Community

1
1

answered Mar 21 '12 at 00:49

Dan Kanze

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Where did the second part of your explanation come from? You're right that `j++ = 1`, so why do you go on to say the `j++ = 2` just because it's followed by a `++j`. – Ken Wayne VanderLinde Mar 21 '12 at 01:05
thanks youre right. i made corrections mate mind lifting that -1? – Dan Kanze Mar 21 '12 at 01:39

Jeff B · Accepted Answer · 2012-03-21T02:10:45.983

I don't think the other answers here are correct.

The variables are evaluated in order according to mathematical ordering operations. In this case, we are just adding, so they are evaluated left to right. j++ and then ++j

j++ + ++j

j let's call the value of j at the start of the loop p
j++ j evaluates to the pre-increment value (p) and then j is incremented (p+1)
++j j is incremented (p+2) and evaluates as the post-increment value (p+2)

So, the two evaluated numbers are p + p+2:

i  initial j  j++ + ++j   k   j
================================
0      1       1  +  3    4   3
1      3       3  +  5    8   5
2      5       5  +  7   12   7
3      7       7  +  9   16   9
4      9       9  + 11   20  11

score 1 · Answer 3 · edited Apr 29 '12 at 20:25

1

j++ increments the value of j, but returns the pre-incremented value.
++j increments the value of j, but returns the incremented value.

edited Apr 29 '12 at 20:25

Charles Menguy

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answered Apr 29 '12 at 13:47

score 1 · Answer 4 · answered Mar 21 '12 at 01:09

In C#, the + operator is just a function, and in your case, j++ and ++j are the arguments to the function. Evaluation of function arguments proceeds from left to right, so here's what we get for each iteration of the loop:

j++ increments the value of j, but returns the original value.
++j increments the value of j again and returns the new value.
The + operator is called with the results from (1) and (2).

For example, when i==0, j is initially 1. Then j++ executes, setting j to 2, and returns 1. Then ++j increments j to 3, and returns 3. So the addition becomes 1 + 3, resulting in k = 4.

why is this the output of the following loop?

4 Answers4