Why does the printf of ++**++argv output 'p'?

Question

#include<stdio.h>
int main(int argc,char *argv[]) {
    printf("%c",++**++argv);
    return 0;
}

Suppose the command line arguments passed were:

./a.out one two three

The output is:

Can someone please explain me what is happening?

The letter `p` is equals to `o + 1`. `one` is the parameter with index 1. — Patrick Bucher, Jan 05 '18 at 11:12
@MichaelWalz True, but maybe this is a riddle to teach the basics? — Patrick Bucher, Jan 05 '18 at 11:12
@PatrickBucher not sure if this is the right way to teach basics. — Jabberwocky, Jan 05 '18 at 11:13
@MichaelWalz teaching this is not teaching basics, it's teaching very bad practices and should not be welcomed on SO, besides, SO is not primarily a site for free tuition. — Martin James, Jan 05 '18 at 11:18
@MartinJames Yes, it's bad practice, and yes SO is not a site for free tuition, but it is still interesting to understand why languages behave they way they do when you abuse them. — klutt, Jan 05 '18 at 11:25
@klutt It might be interesting to some, but it's not generally useful to future SO users/visitors. There are some homework/quiz questions that can be interesting AND useful. This is not one of them. — Martin James, Jan 05 '18 at 11:34
@MichaelWalz I think this snippet is a coding puzzle. It has no practical uses aside from providing a somewhat nerdy entertainment to its readers. — Sergey Kalinichenko, Jan 05 '18 at 11:38
It is perfectly clear what the question is asking. Putting it on hold as "unclear" makes zero sense to me. Voting to re-open. — Sergey Kalinichenko, Jan 05 '18 at 17:23

score 6 · Answer 1 · answered Jan 05 '18 at 11:17

Start from the back of the ++**++argv expression:

argv starts off as a pointer to element zero, i.e. "./a.out" or ""
++argv is char** that points to string "one"
*++argv is char* that points to string "one"'s initial element
**++argv is char that is equal to string "one"s initial element, i.e. 'o'
++**++argv is char that follows 'o'. On most systems that's 'p'.

The last operation modifies program's arguments in place, which is allowed by the standard (Q&A).

1 Answers1