|
This code is intended to print 20 minus signs, but it does not work:
#include <stdio.h>
main()
{
int i = 0, n = 20;
do
{
printf("-");
i--;
} while ( i < n );
}
There are at least 4 ways to fix it by changing exactly one character. Find!
|
|
|
|
|
≡
|
2009 Jul 1 at 23:30 UTC
— Ed. 2009 Jul 5 at 20:14 UTC
|
|
|
Down Rodeo
Cap'n Moth of the Firehouse
2007 Oct 19 • 5486
57,583 ₧
|
Change the i-- to n--? I don't know why that's a question, it works. But it should return an int :p
Not so sure what the others are; I shall have a little more think-time.
|
|
|
|
|
≡
|
2009 Jul 2 at 00:09 UTC
|
|
|
|
*** You have found 1 of the 4 solutions. ***
This is just a puzzle, don't worry about compiler warnings, etc. You don't have to return an int to get it to compile, anyway.
|
|
|
|
|
≡
|
2009 Jul 2 at 00:18 UTC
|
|
|
Down Rodeo
Cap'n Moth of the Firehouse
2007 Oct 19 • 5486
57,583 ₧
|
Ya, it compiled warning-free, which surprised me. I expected it to complain about my lack of a newline but nope, nothing.
|
|
|
|
|
≡
|
2009 Jul 2 at 00:25 UTC
|
|
|
|
...
} while ( -i < n );
Maybe?
|
|
|
|
|
≡
|
2009 Jul 2 at 20:18 UTC
|
|
|
Down Rodeo
Cap'n Moth of the Firehouse
2007 Oct 19 • 5486
57,583 ₧
|
I was going to ask if changing whitespace was allowed but that looks like a solution, kd.
|
|
|
|
|
≡
|
2009 Jul 3 at 21:15 UTC
|
|
|
|
*** You have found 2 of the 4 solutions. ***
Good job killer.
Yes, you can change a white-space character. But only ONE!
|
|
|
|
|
≡
|
2009 Jul 4 at 00:29 UTC
|
|
|
|
ok im just gona ask
what
how
txt doc?
i dont know
can you explane
I drink to forget but I always remember.
|
|
|
|
|
≡
|
2009 Jul 4 at 01:13 UTC
|
|
|
Down Rodeo
Cap'n Moth of the Firehouse
2007 Oct 19 • 5486
57,583 ₧
|
It helps if you understand the basics of imperative programming. You don't even really need a compiler, thinking through it ought to work.
Uhm, changing the assignment n =-20; might work?
|
|
|
|
|
≡
|
2009 Jul 5 at 11:00 UTC
|
|
|
|
Down Rodeo said: Uhm, changing the assignment n =-20; might work?
that will only print one minus sign because i is never less than n.
|
|
|
|
|
≡
|
2009 Jul 5 at 19:50 UTC
|
|
|
Down Rodeo
Cap'n Moth of the Firehouse
2007 Oct 19 • 5486
57,583 ₧
|
This is true. I didn't check it, I was guessing 
|
|
|
|
|
≡
|
2009 Jul 6 at 16:44 UTC
|
|
|
|
WOW the most funny and newbie c puzzle ever...the increment shuld be
i++;
or make
.
.
.
int i=20,n=0;
do{
.
.
}while(i>n);
|
|
|
|
|
≡
|
2009 Jul 9 at 15:18 UTC
— Ed. 2009 Jul 9 at 15:19 UTC
|
|
|
Down Rodeo
Cap'n Moth of the Firehouse
2007 Oct 19 • 5486
57,583 ₧
|
But you see, thenotsogreat, neither of those involves changing exactly one character, which means you haven't solved the puzzle, friend.
|
|
|
|
|
≡
|
2009 Jul 9 at 16:10 UTC
|
|
|
|
I don't remember how exactly the while check works (is 0 true or false?), but would this work?
Anyway just bumping this truck.
...and that's the bottom line because Mate de Vita said so.
|
|
|
|
|
≡
|
2010 Jan 16 at 15:48 UTC
— Ed. 2010 Jan 16 at 15:49 UTC
|
|
|
Down Rodeo
Cap'n Moth of the Firehouse
2007 Oct 19 • 5486
57,583 ₧
|
Ah, clever. Well done :) zero evaluates to false, all other values are true, I think.
|
|
|
|
|
≡
|
2010 Jan 16 at 17:28 UTC
|
|
|
|
It took a while but you got it. GG Mate Davate.
|
|
|
|
|
≡
|
2010 Jan 23 at 02:28 UTC
|
|
|
Down Rodeo
Cap'n Moth of the Firehouse
2007 Oct 19 • 5486
57,583 ₧
|
So you've resparked my interest (at 3 am, you bas), and I'm not sure I'll ever get the answer. Because: can't change i--; one change is a solution and anything else is not an operator. Can't change the assignment, that's a two-character change. Can't change the assignment of i to anything useful (largest is 9). Can't change the type of loop (obviously). Can't change the loop boundaries (not that I think it would help). Can't change the thing it's printing to anything useful.
Ah... 
Legit, you made me break out the pens. I want to say that changing the < to & will work, but it might actually print out 19 or 21 minus signs, I think. It's late! We want the pattern
code 0 0 0 1 0 1 0 0 & 1 1 1 0 1 0 1 1 That means i = -21, so there should be 21 minus signs printed. I am probably wrong somewhere so please let me know!
|
|
|
|
|
≡
|
2012 Apr 3 at 02:34 UTC
— Ed. 2012 Apr 3 at 02:35 UTC
|
|
|
Down Rodeo
Cap'n Moth of the Firehouse
2007 Oct 19 • 5486
57,583 ₧
|
Since you said you need the attns!
|
|
|
|
|
≡
|
2012 May 22 at 23:14 UTC
|
|
|
|
You're right that it prints 21 minus signs.
Are you really not just running it to see??
shell code jwilson@snickerdoodle ~$ cat >puz.c #include <stdio.h> main() { int i = 0, n = 20; do { printf("-"); i--; } while ( i & n ); } ^D jwilson@snickerdoodle ~$ make puz cc puz.c -o puz jwilson@snickerdoodle ~$ ./puz | wc 0 1 21 |
|
|
|
|
≡
|
2012 May 24 at 23:57 UTC
|
|
|
|
It's only 30-odd keystrokes!
cat >puz.c^J^V^Dmake puz^J./puz | wc^J
|
|
|
|
|
≡
|
2012 May 24 at 23:59 UTC
|
|
|
|
|
Down Rodeo
Cap'n Moth of the Firehouse
2007 Oct 19 • 5486
57,583 ₧
|
I'm on Windows when I visit your site. Sadly, my laptop is poorly right now, as it has a case of the dead fan. So no I'm really not running the code :p
|
|
|
|
|
≡
|
2012 May 26 at 01:02 UTC
|
|
|
|
That's what IDEONE is for!
|
|
|
|
|
≡
|
2012 May 30 at 22:50 UTC
|
|
|
|
I went through multiple bitwise operators, such as left shift (infinite loop) - you cannot do right shift without changing more than one character, bitwise or and xor (infinite loop and no output), and even tried the unary operator ~ (which, like DR's solution, prints 21 minuses).
But in the end, this: appears to print 20 minuses.
...and that's the bottom line because Mate de Vita said so.
|
|
|
|
|
≡
|
2012 Jun 23 at 12:54 UTC
— Ed. 2012 Jun 23 at 13:02 UTC
|
|
|
|
By jove I think you've got it!
|
|
|
|
|
≡
|
2012 Jun 23 at 21:33 UTC
|
|
|
|
Page [1]
|
