Number rounding

[xxpokerxx]

Hi, is it possible to specify rounding to a certain position.

Basically i need to :

given this float what is it (hundreds, thousands, tens of thousands, millions, etc.) and then round up to the nearest whole number in that position.

So,

given 17,029 round to 18,000
given 1,348 round to 2,000
given 1,564,783 round to 2,000,000
given 443,756 round to 500,000

if it is possible, could anyone point me in the right direction. Thanks.

[rcirka]

You may want to try NSNumberFormatter. It gives various options for rounding. I haven't tried it for non-decimal rounding but it may work for you.

[nobre84]

A pretty simple idea would be , keep dividing your number by 10 and place in a variable the total amount divided by, until your make the number < 10. Then, round up this number and multiply it by the total amount you divided earlier.

edit: this algorithm will round 17029 to 20000. The others will work as you said.

[xxpokerxx]

Thanks for the responses, i got it. I actually changed what i wanted a little bit, but its the same idea. instead of just rounding up the largeest digit, it rounds the second largest (the values are for use in a graph so i only need a small gap between the rounds. If I chose 1,100,000 with my previous request i would get 2,000,000 which is a huge difference and not good, but now i will get 1,200,000 which works out perfectly.

My solution, not pretty but it works:

NSNumberFormatter *formatter = [[NSNumberFormatter alloc] init] autorelease];
[formatter setRoundingMode:NSNumberFormatterRoundCeiling];

float digitToRound = pow(10, (double)[f stringFromNumber:[NSNumberWithFloat:floatToRound] length] - 2]);

return [[formatter stringFromNumber:[NSNumber numberWithFloat:(floatToRound / digitToRound)]] floatValue] * digitToRound;

this stops rounding when floatToRound is double digits and returns NAN when floatToRound is single digits. however for my purposes, those cases will never occur so i dont care.