I met a trouble about float.Help me

Question

float abc = 20130401.0f;
NSLog(@"abc = %f",abc);

result:

2013-04-08 15:47:38.963 CCLearnIphone[5034:c07] abc = 20130400.000000

It's the same to use double.

How can I express an eight-digit number?

I know precision is lost but .000000 should not be 1.0000000.

%f should be 0.9xxxxx etc

why is it .000000 rounded to 1.000000?

If you need greater precision use a `double`. You can see a similar answer here: http://stackoverflow.com/questions/5098558/float-vs-double-precision — Alladinian, Apr 08 '13 at 08:49
@Alladinian It's the same to use double.How can I express an eight-digit number? — ccimorningbell, Apr 11 '13 at 03:24

Mirko Catalano · Answer 1 · 2013-10-30T01:20:48.100

0

your number is too hight for a float var so you have to change it like double and remove the 'f' character at the end of your number

double abc = 20130401.0;
NSLog(@"abc = %f",abc);

edited Oct 30 '13 at 01:20

answered Oct 28 '13 at 22:27

Mirko Catalano

score -1 · Answer 2 · edited May 23 '17 at 12:21

-1

It's due to the precision of float: about 7 decimal digits.

edited May 23 '17 at 12:21

Community

answered Apr 08 '13 at 08:49

Nikolai Ruhe

score -1 · Answer 3 · answered Apr 08 '13 at 08:59

-1

why is it .000000 rounded to 1.000000?

That's not the case. 20130401 is rounded to 20130400 because 20130400 is the closest value that a float can represent.

answered Apr 08 '13 at 08:59

Nikolai Ruhe

why `float abc = 19999991`, prints to 19999992. Any Hard and fast rule, so that without computer we can tell the result? – Anoop Vaidya Apr 08 '13 at 09:02
@AnoopVaidya Without calculating exponent and mantissa it's difficult to tell if a value is exactly representable in a floating point type. The simplest rule of thumb is: Less than 7 decimal digits. – Nikolai Ruhe Apr 08 '13 at 09:06
Here's an IEEE 754 converter that shows binary and decimal representations of any entered number: http://www.h-schmidt.net/FloatConverter/ – Nikolai Ruhe Apr 08 '13 at 09:09

3 Answers3