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January 25th, 2004 at 6:16 am
Hmm. Did a search and found this. http://mindprod.com/jgloss/widthindigits.html
If you are strictly dealing with ints, it seems like this method is the most optimized as it doesn’t require any method calls or logorithmic computation. It performs at most 3 simple comparisons.
/**
* Counts number decimal digits in a 32 bit signed number. 0 => 1, 9 => 1,
* 99 => 2, 2,147,483,647 => 10
*
* @param x
* number whose digits you wish to count. Must lie in range 0 ..
* Integer.MAX_VALUE;
*
* @return number of digits in x, e.g. Integer.toString(x).length()
* @author Marc Chappuis marnic@ludomedia.ch
*/
public static int widthInDigits(int x) {
// do an unravelled binary search
if (x
April 27th, 2009 at 8:06 am
int n = 12345;
String s = String.valueOf(n);
int counted = s.length();
July 23rd, 2009 at 8:53 pm
@Min the issue with this approach is you have to explicitly set the bounds. This is fine for base 10, but what about base 2? One can represent a pretty long binary number in a “long” datatype.
@Absar String operations are much expensive than numerical operations. Also, this does not take into account other bases (though it could, by using a STRING->NBASE function).
September 17th, 2009 at 1:09 am
It doesnt seem to work when I enter a 7 digit number, I get a cound of 6.
It works on 6 digits and 8 digits but not 7. Probably more errors with other values as well..