Formatting a (large) number "12345" to "12,345" - iphone

Question

Say I have a large number (integer or float) like 12345 and I want it to look like 12,345.
How would I accomplish that?
I'm trying to do this for an iPhone app, so something in Objective-C or C would be nice.

Answer 1

Here is the answer.
NSNumber* number = [NSNumber numberWithDouble:10000000];
NSNumberFormatter *numberFormatter = [[NSNumberFormatter alloc] init];
[numberFormatter setNumberStyle:kCFNumberFormatterDecimalStyle];
[numberFormatter setGroupingSeparator:#","];
NSString* commaString = [numberFormatter stringForObjectValue:number];
[numberFormatter release];
NSLog(#"%# -> %#", number, commaString);

Answer 2

Try using an NSNumberFormatter.
This should allow you to handle this correctly on an iPhone. Make sure you use the 10.4+ style, though. From that page:
"iPhone OS: The v10.0 compatibility mode is not available on iPhone OS—only the 10.4 mode is available."

Answer 3

At least on Mac OS X, you can just use the "'" string formatter with printf(3).
$ man 3 printf
`'' Decimal conversions (d, u, or i) or the integral portion
of a floating point conversion (f or F) should be
grouped and separated by thousands using the non-mone-
tary separator returned by localeconv(3).
as in printf("%'6d",1000000);

Answer 4

Cleaner C code
// write integer value in ASCII into buf of size bufSize, inserting commas at tousands
// character string in buf is terminated by 0.
// return length of character string or bufSize+1 if buf is too small.
size_t int2str( char *buf, size_t bufSize, int val )
{
char *p;
size_t len, neg;
// handle easy case of value 0 first
if( val == 0 )
{
a[0] = '0';
a[1] = '\0';
return 1;
}
// extract sign of value and set val to absolute value
if( val < 0 )
{
val = -val;
neg = 1;
}
else
neg = 0;
// initialize encoding
p = buf + bufSize;
*--p = '\0';
len = 1;
// while the buffer is not yet full
while( len < bufSize )
{
// put front next digit
*--p = '0' + val % 10;
val /= 10;
++len;
// if the value has become 0 we are done
if( val == 0 )
break;
// increment length and if it's a multiple of 3 put front a comma
if( (len % 3) == 0 )
*--p = ',';
}
// if buffer is too small return bufSize +1
if( len == bufSize && (val > 0 || neg == 1) )
return bufSize + 1;
// add negative sign if required
if( neg == 1 )
{
*--p = '-';
++len;
}
// move string to front of buffer if required
if( p != buf )
while( *buf++ = *p++ );
// return encoded string length not including \0
return len-1;
}

Answer 5

I did this for an iPhone game recently. I was using the built-in LCD font, which is a monospaced font. I formatted the numbers, ignoring the commas, then stuck the commas in afterward. (The way calculators do it, where the comma is not considered a character.)
Check out the screenshots at RetroJuJu. Sorry--they aren't full-sized screenshots so you'll have to squint!

Answer 6

Hope that helps you (it's in C) :
char* intToFormat(int a)
{
int nb = 0;
int i = 1;
char* res;
res = (char*)malloc(12*sizeof(char));
// Should be enough to get you in the billions. Get it higher if you need
// to use bigger numbers.
while(a > 0)
{
if( nb > 3 && nb%3 == 0)
res[nb++] = ',';
// Get the code for the '0' char and add it the position of the
// number to add (ex: '0' + 5 = '5')
res[nb] = '0' + a%10;
nb++;
a /= 10;
}
reverse(&res);
return res;
}
There might be a few errors I didn't see (I'm blind when it comes to this...)
It's like an enhanced iToA so maybe it's not the best solution.

Answer 7

Use recursion, Luke:
#include <stdio.h>
#include <stdlib.h>
static int sprint64u( char* buffer, unsigned __int64 x) {
unsigned __int64 quot = x / 1000;
int chars_written;
if ( quot != 0) {
chars_written = sprint64u( buffer, quot);
chars_written += sprintf( buffer + chars_written, ".%03u", ( unsigned int)( x % 1000));
}
else {
chars_written = sprintf( buffer, "%u", ( unsigned int)( x % 1000));
}
return chars_written;
}
int main( void) {
char buffer[ 32];
sprint64u( buffer, 0x100000000ULL);
puts( buffer);
return EXIT_SUCCESS;
}