Re:How many Java programmers does it take to round up to a p...
by tmm1 on Friday November 02, 2007 @12:14AM
(#75321)
http://pastie.caboo.se/113312
[ Parent
| on HN ]
Re:How many Java programmers does it take to round up to a p...
by tallpapab on Friday November 02, 2007 @03:04PM
(#75542)
You have an infinite loop when i < 2^30.
static int nextOne(int theInt) {
int answer = 1;
while (answer < theInt && 0 < answer) {
answer <<= 1;
}
if (answer < 1) {
throw new IllegalArgumentException("Sorry, " +
theInt + " is bigger than 2 to the power of 30 (1073741824).");
}
return answer;
}
Re:How many Java programmers does it take to round up to a p...
by gojomo on Friday November 02, 2007 @04:14PM
(#75573)
Good point. (And, as your signature indicates, this is more likely to be a static utility method.)
If I were defensively programming against unanswerable input, I'd tend towards failing fast with a single early check:
static int roundUpToPowerOfTwo(int i) {
if(i>(1<<30)) throw new IllegalArgumentException("Answer overflows int");
int rounded = 1;
while(rounded<i) rounded <<= 1;
return rounded;
}
If we needed to give answers for any size integer (or at least up to 2^(2^31-1)), we could use BigInteger, whose #bitLength method is useful: static BigInteger roundUpToPowerOfTwo(BigInteger i) {
return BigInteger.ONE.shiftLeft(i.subtract(BigInteger.ONE).bitLength());
}
And if we weren't being literal-minded about the original request, it might make sense to just return the exponent rather than the power-of-2 itself, and let the caller worry about overflows: static int ceilLog2(long val) {
return BigInteger.valueOf(val-1).bitLength();
}
Of course we are ignoring negative exponents, in the integer spirit of the original question.There, is it done to death yet? :)
Re:How many Java programmers does it take to round up to a p...
by dhouston on Friday November 02, 2007 @04:16AM
(#75382)
float precision issues make me nervous (although it's probably ok here), not to mention log() is kind of expensive compared to bit shifting
why not just something like
def fn(n): ret=1; while ret < n: ret <<= 1; return ret
(news.yc/browsers mangle whitespace; proper formatting is left as an exercise to the reader)
Re:How many Java programmers does it take to round up to a p...
by Goladus on Friday November 02, 2007 @07:53AM
(#75406)
To format code, add a whole bunch of returns and then indent 4-5 spaces
(5 returns and 6 spaces later
whitespace does what I want)
I think it's supposed to be 4 and 4, but I had some issues once and since then I just pile on returns and spaces until I get bored.
Re:How many Java programmers does it take to round up to a p...
by nostrademons on Monday November 05, 2007 @09:59AM
(#76277)
I thought it's 2 and 1:
print 'Hello world'
Yup, that's it. A blank line before the indented block, then any amount of indentation. Reddit is 2 and 4.
Re:How many Java programmers does it take to round up to a p...
by gojomo on Friday November 02, 2007 @04:42AM
(#75387)
You could use that approach in Java...
public int roundUpToPowerOfTwo(int i) {
return 1 << (int)(Math.ceil(Math.log(i)/Math.log(2)));
}
...but the first approach that occurred to this Java coder was instead... public int roundUpToPowerOfTwo(int i) {
int rounded = 1;
while(rounded<i) rounded <<= 1;
return rounded;
}