[ipsych]

Please understand that I am working on my very first Cocoa application, so I certainly may just be missing something obvious.

In my application I have an NSTextField which displays the float value result of a calculation. I'd like to have the value displayed rounded to the nearest .5. For instance, 2.1 would display as 2 and 2.3 would display as 2.5. I tried the following, but it didn't work:

NSNumber * roundingIncrement = [NSNumber numberWithFloat: .5];
NSNumberFormatter * myFormatter = [[[NSNumberFormatter alloc] init] autorelease];

[myFormatter setRoundingIncrement: roundingIncrement];
// Insert other setters for minimum, maximum, format strings, etc.
[myTextField setFormatter: myFormatter];

Any help or insights will be appreciated. Thanks!

[techtrucker]

Normally I would turn to plain old C in a case like this, but rounding in C has always been ugly at best ( http://groups.google.com/group/comp....rch+this+group )

You could look into:

- (NSDecimalNumber *)decimalNumberByRoundingAccordingToBehaviorid <NSDecimalNumberBehaviors>)behavior

I'm sure others more familiar with floating point math will chime in here. Good luck!

[Amorya]

double the number, add 0.5, take the integer part then half it?

2.1*2=4.2
4.2+0.5=4.7
int part = 4
halved = 2

2.3*2+0.5=5.1
int part = 5
halved = 2.5


Amorya

[ipsych]

Thanks, Amorya! That did the trick. I generalized it below to work to find the nearest .25, .5, .75, etc.

float decimalValue;
float decimalIncrement = .25; // Set to .25, .5, .75, etc.

decimalValue = decimalValue * (1/decimalIncrement);
decimalValue += decimalIncrement;
decimalValue = (int)decimalValue * decimalIncrement;