How do I round off a Float value in C

Question

I'm looking for function which can Round off a float value in C ,Suppose have number 0.1153846 should be rounded off till 6 decimal producing output as 0.115385

Though there is function in objective c like lroundf() but not sure how can use it in my context.

I'm on gcc compiler and any help would be much appreciated.

possible duplicate of [round() for float in C++](http://stackoverflow.com/questions/485525/round-for-float-in-c) — Tony The Lion, Sep 14 '12 at 09:25
Not a good duplicate: This post is interested in rounding to a number of decimal places (6). Suggested [dupe](http://stackoverflow.com/questions/485525/round-for-float-in-c) is for rounding to a whole number. There are subtle and important different issues involved. — chux - Reinstate Monica, Jan 29 '16 at 18:08

score 1 · Answer 1 · answered Sep 14 '12 at 09:32

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float f = 0.1153846;
f = floor(f * 1000000) / 1000000;

This should work.

answered Sep 14 '12 at 09:32

User 99x

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This is something I also come up with. let me check how its working – Amit Singh Tomar Sep 14 '12 at 09:41
It alone won't work guess something like f+0.5 needs to be done!! – Amit Singh Tomar Sep 14 '12 at 09:43
@KerrekSB +5 should be added as you mentioned to round up to the next digit. thanks. – User 99x Sep 14 '12 at 09:51
f = floor((f * 1000000)+0.5) / 1000000; would be ok – Amit Singh Tomar Sep 14 '12 at 09:56
@AmitSinghTomar: Yeah, that's the one :-) – Kerrek SB Sep 14 '12 at 10:05
@AmitSinghTomar `floor((f * 1000000)+0.5) / 1000000;` has a number of problems 1) Negative numbers, 2) values when `f * 1000000` ULP is 0.5 or 1.0, 3) value when `f * 1000000` is just under 0.5. – chux - Reinstate Monica Aug 21 '15 at 19:35

Basile Starynkevitch · Answer 2 · 2012-09-14T09:56:39.287

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You might want to do

  double x = 0.1153846;
  double rx = round (x * 1e6) * 1.e-6;

However, remember that IEEE 754 floating points are binary, with base 2 mantissa.

edited Sep 14 '12 at 09:56

answered Sep 14 '12 at 09:33

Basile Starynkevitch

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but output comes to be 0.115380 which is not what I'm looking for. – Amit Singh Tomar Sep 14 '12 at 09:46
Try with `1e6` and `1.e-6` instead – Basile Starynkevitch Sep 14 '12 at 09:57
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A slightly more pedantic answer would use `/` as in `double rx = round (x * 1e6) /1.e6;` – chux - Reinstate Monica Aug 21 '15 at 19:31

AProgrammer · Answer 3 · 2012-09-14T10:46:33.473

It's quite unusual for float to be a decimal type, that means that whatever you do, the result of the rounding will most often than not not be representable in a float and will be adjusted again to match a representable number.

If such a rounding is really needed for computational purpose, a floating point type is probably not the correct type to use.

If it is just for display purpose, use the control the printf family give you.

score 1 · Answer 4 · edited Aug 21 '15 at 18:59

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int precision = 3;
float num = 1.63322;
printf("%.*f",precision,num);

edited Aug 21 '15 at 18:59

josliber

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answered Aug 21 '15 at 18:37

aman

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This does not round `num`. It does print a rounded version, but `num` is not affected. – chux - Reinstate Monica Jan 29 '16 at 18:11

How do I round off a Float value in C

4 Answers4