numbers

I have a number formatter set up to convert currency strings to decimal values. The problem is that if the text string does not have a leading dollar sign ("$"), it gets converted to 0, rather than a valid matching number. So: "$3.50" converts to 3.50 "3.50" converts to 0 Here is the code for the converter: // formatter to convert a value to and from a currency string NSNumberFormatter *currencyFormatter = [[NSNumberFormatter alloc] init]; [currencyFormatter setNumberStyle:NSNumberFormatterCurrencyStyle]; [currencyFormatter setGeneratesDecimalNumbers:YES]; Am I missing something? I ran into a

Is there somewhere hidden in a (small) module a function, which does this for me: my $var = 23654325432; $var = reverse $var; $var =~ s/\d{3}\K(?=\d+)/_/g; $var = reverse $var; I like Number::Format but it didn't pass all tests on windows. use Number::Format; my $nf = new Number::Format( -thousands_sep => ',', -decimal_point => '.', ); my $formatted = $nf->format_number( 23142153 ); JRFerguson Look at perl5i . It has commify and group_digits methods. The fastest method is in the perlfaq ( perldoc -q commas ): sub commify { local $_ = shift; 1 while s/^([-+]?\d+)(\d{3})/$1,$2/; return $_; } Of

Is there any published O(log b) algorithm to determine if a b-bit number is square of an integer? (I apologize if this question is beyond the scope of this site and would happily retrieve it if so) Update: I realize that my question as posed is unreasonable. So let me amend it by asking for any algorithm that is sub polynomial operations in b. Not necessarily O(log^k b) for a constant k, and has "reasonable" space complexity. Operations are defined in the usual sense: something that is reasonable for the task at hand (e.g. add, negate, xor, and, or, etc.) Post script : Now I realize that my

I am looking at solution for the set bit count problem (given a binary number, how to efficiently count how many bits are set). Here, http://graphics.stanford.edu/~seander/bithacks.html#CountBitsSetNaive , I have found some methods. What about the lookup table method? I dont understand what properties of binary representation / number make it work. static const unsigned char BitsSetTable256[256] = { # define B2(n) n, n+1, n+1, n+2 # define B4(n) B2(n), B2(n+1), B2(n+1), B2(n+2) # define B6(n) B4(n), B4(n+1), B4(n+1), B4(n+2) B6(0), B6(1), B6(1), B6(2) }; unsigned int v; // count the number of

I'm working on getting an old piece of code working, from 2003. I'm trying to replicate an ATM style decimal textbox. This code claims to have worked for someone, but I am having trouble implementing it. Maybe someone has a better way of achieving this? Maybe in jQuery? This is how I would solve it: http://jsfiddle.net/77bMx/86/ Handles numpad input as well as standard number keys Works with backspace Will revert to 0.00 if you somehow manage to produce illegal input (like backspacing too much) The basic idea is to intercept any input to the box, make sure it's of the correct type (a number,

I am trying to find a way to reverse a number without Converting it to a string to find the length Reversing the string and parsing it back Running a separate loop to compute the Length i am currently doing it this way public static int getReverse(int num){ int revnum =0; for( int i = Integer.toString(num).length() - 1 ; num>0 ; i-- ){ revnum += num % 10 * Math.pow( 10 , i ); num /= 10; } return revnum; } But I would Like to implement the above 3 conditions. I am looking for a way , possibly using the bit wise shift operators or some other kind of bitwise operation. Is it possible ? If so how

I'm writing convenience methods to check if number is positive or negative like so: class Numeric def positive? self > 0 end def negative? self < 0 end end but in this case I do not know how to handle cases like these: >> 0.positive? >> 0.negative? Update : I've updated the typo in the class name. I used numeric because I needed to check the floats as well. If the problem is that you're getting false for both, either you consider 0 to be positive or not. If so, you should have something like: def positive? self >= 0 end If not, leave it as it is, since 0 is neither positive not negative and

How can I merge two integers from a list into one? (in Scheme) Example: '(11 223) -> 11223 Assuming that the list has exactly two elements, and that both are numbers: (define (merge-numbers lst) (let ((1st (number->string (first lst))) (2nd (number->string (second lst)))) (string->number (string-append 1st 2nd)))) It works as expected: (merge-numbers '(11 223)) > 11223 Alternatively, without using a let : (define (merge-numbers lst) (string->number (string-append (number->string (first lst)) (number->string (second lst))))) This is my original answer from Jan 25 '12 at 3:05. It only handles