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user:1234 user:me (yours) |
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score:3 (3+) score:0 (none) |
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answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
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A popularity-contest is a competition where the answer with the highest vote tally (upvotes minus downvotes) wins. As these are frequently closed, read the tag info and post your challenge to the sandbox first.
192
votes
80
answers
26k
views
What? No error? [closed]
Your task is simple. Write a program that should obviously produce an error on first glance either when compiled or run, but either doesn't or produces some other unrelated error. This is a popularity …
52
votes
Make it look like I'm working
Lets go with a simple bash script that makes you look hackerish by printing the contents of every file identified as text in /var/log/ line by line, with random delays to make it look like intensive t …
20
votes
Weirdest obfuscated "Hello World!"
Brainf***
Not sure if this really counts as obfuscated, but it never uses more than four plus or minus symbols in a row. So there's that.
>>+++[<+++>-]<+[>+++>>+++[<+++>-]<+[<]>-]<+++[>+++<-]>+[>+++ …
11
votes
20
answers
1k
views
Every 2^n times
Let n be the number of times your program has been executed. If n is a power of 2, then print 2^x where n = 2^x; otherwise, simply output the number. Example run:
[1st time] 2^0
[2nd time] 2^1
[3rd t …
2
votes
Most creative way to display 42
F#
let answer = 42
[|62; 74; 74; 70; 73; 16; 5; 5; 77; 77; 77; 4; 61; 69; 69; 61; 66; 59; 4; 57;
69; 67; 5; -7; 71; 19; 74; 62; 59; 1; 55; 68; 73; 77; 59; 72; 1; 74; 69; 1;
66; 63; 60; 59; 1; 74; …
1
vote
Every 2^n times
Ruby
Alright, I think I'll try this now. It searches itself for the definition of n.
def p2 n
n == 1 ? 0 : p2(n >> 1) + 1
end
n = 1
if (n != 0) & (n & (n - 1) == 0) || n == 1
puts("2^" + (p2(n). …
1
vote
Random script that isn't actually random
F#
This solution is very flexible -- it allows you to pick people in batches, so you won't have to run it again for a little while. Supply the number as the first argument to the program:
[<EntryPoi …
0
votes
Write a function that takes (x, y) and return x to the power of y WITHOUT Loops
F#
Since ^ and ** were already taken and infix operators are fun, let's call it ^^
let rec (^^) (x: int) y =
let fx = float x
match y with
| y when y < 0 -> (x ^^ y + 1) / fx
| 0 -> 1.0
| …
-5
votes
5
answers
458
views
What time is it? [closed]
I really hate looking up to the top right of the screen to look at the time on OS X's clock widget; I'd much rather keep my eyes at least somewhere near my workspace to check the time. I also hate oth …