unexpected result from printf character print

Question

#include<stdio.h>
main()
{
  printf("%c\n",1-3+2["nexus"]);
}

The result is v. How does it turn out?
What does the indentation (square ones) do?

What did you expect? Please also reduce your example, as you probably don't need the `1-3+` part to affect the expectations. — Cœur, Mar 13 '18 at 05:23
@Cœur, just to let you know, I didn't DV your last question. I just marked it. (I don't dv, I just try and help) — David C. Rankin, Mar 13 '18 at 07:20
@DavidC.Rankin I appreciate your help. But I feel bad about the downvotes I got after all the time I spent searching for "square brackets syntax" and "square brackets notation". — Cœur, Mar 13 '18 at 07:52
Yes, just shake them off, they come out in the wash. It used to get under my skin when the downvote weenies would run by without having the integrity to leave a comment, but in the end, it's just 2-points, and there are some people that just feel superior by being able to hit the down button. Just make sure you set a better example. That's all we can do. (and things have gotten better -- it now costs the downvoter 1-point to downvote, so they are not nearly as willy nilly as they were in the past) — David C. Rankin, Mar 13 '18 at 07:57

score 0 · Answer 1 · answered Apr 06 '14 at 19:22

2["nexus"] is probably leading us to 3 element of the nexus considering it as an array of characters[arr[2]=x in our case]

following is 1-3viz -2 added to x character ie added to to ascii value and the corresponding element is v

cant exactly figure out what the square identation really mean but according to the oput this is possible way .

score 0 · Answer 2 · answered Mar 13 '18 at 05:38

2["nexus"] is same as "nexus"[2], which correspond to the 2nd index or the 3rd element of this string. ( Which is 'x' here ).

In equation, 1-3+ 2["nexux"] is like 1-3+'x' that represent 'v'.

Note

When a string is declared like "nexus".

nexus[0] is 'n'
nexus[1] is 'e'
nexus[2] is 'x'
nexus[3] is 'u'
nexus[4] is 's'

2 Answers2