Add zeros in the right hand of a float number

Question

I need to do this (does not round, floor or ceil):

example:

1.58700023 (16bits)

expected 1.58700000

i am doing this operation: (value-32768.0)/32768 to convert to (byte value to real value in float) int his conversion have this error = X.00000023 or others

Possible duplicate of [Is floating point math broken?](https://stackoverflow.com/questions/588004/is-floating-point-math-broken) — too honest for this site, Jun 16 '17 at 08:48

score 1 · Answer 1 · answered Jun 16 '17 at 08:40

1

It's quite possible that you cannot do this with floats; note that not all values are exactly representable as a float: it's limited to 32 (typically) bits after all.

For printing, you should be able to use:

printf("%.3f", 1.58700023);

This will print 1.587 by rounding as the value is converted to string.

answered Jun 16 '17 at 08:40

unwind

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It would be also a nice idea to point out how comparison can be fixed (just like you proposed a fix for printing. Because that I guess must be an issue. – Ajay Brahmakshatriya Jun 16 '17 at 09:19
dont work for me: result 1.58700023 1.58799997 EDIT: i am using float – Jesús Álvarez Jun 16 '17 at 09:20

csl · Accepted Answer · 2017-06-16T08:49:02.297

I assume you want to implement rounding with primitives.

To round to four decimal places, multiply with 10e4, convert to integer (effectively truncating or removing the decimals), then divide by float value 10e4 which converts the result back to float again. For example:

#include <stdio.h>

int main()
{
  double f = 1.58700023;
  printf("%10.10f\n", f);

  // Round to 4 decimal places
  double r = (int)(f*10000.0)/10000.0;
  printf("%10.10f\n", r);

  return 0;
}

This outputs:

1.5870002300
1.5870000000

There are many edge cases not supported by this routine, though. E.g., you may want to add one half of 1/10e4 to perform rounding to nearest next digit, etc.

Add zeros in the right hand of a float number

2 Answers2