Round up a float number that has a non-zero decimal digit

Question

I was asking about round a number half up earlier today and got great help from @alk. In that post, my thinking was to round up 4.5 to 5 but round 4.4 down to 4. And the solution given by @alk was:

int round_number(float x)
{
 return x + 0.5;
}

and it works very elegantly!

In this post, I would like to discuss how to implement the ceil() function in C. Along the same line as the last solution given by @alk, I came up with the following:

int round_up(float y)
{
   return y + 0.99999999;
}

This works for all situations except when the the float number y has .00000001. I am wondering if there's any better way to do the same thing as ceil() in C.

You question asked how to **round up** then in the content you said **ceil()** function. That's completely different. You should edit it — phuclv, Mar 02 '14 at 07:16
I don't think there's a way to do this reliably without getting in bed with the floating point format. — Hot Licks, Mar 03 '14 at 00:03

Mike Woolf · Answer 1 · 2014-03-02T05:33:48.603

2

Unless you reliably know the epsilon of float (I'm not sure standard C provides that), I think you're stuck with return (y < 0 || y == (int)y) ? y : y + 1;

edited Mar 02 '14 at 05:33

answered Mar 02 '14 at 04:42

Mike Woolf

1,210
7
11

Good catch; I've modified slightly to handle negative values. – Mike Woolf Mar 02 '14 at 05:34

chux - Reinstate Monica · Answer 2 · 2014-03-03T00:35:03.707

This fails for negative numbers.

int round_up(float y) {
 return y + 0.99999999;
}

But let's use that to our advantage. float to int conversion is a truncate toward 0.0. Thus negative numbers are doing a "round up" or "ceiling" function. When we have a positive float, convert to int noting this is a "floor" function. Adjust when y is not an integer.

(Assume y within INT_MIN ... INT_MAX.)

int ceil(float y) {
   if (y < 0) {
     return y;  // this does a ceiling function as y < 0.
   }
   int i = y; // this does a floor function as y >= 0.
   if (i != y) i++;
   return i;
 }

void ceil_test(float y) {
   printf("%f %d\n", y, ceil(y));
}

@Lưu Vĩnh Phúc The ceiling of -2.5 is -2, not -1. – chux - Reinstate Monica Mar 02 '14 at 17:54 — chux - Reinstate Monica, Mar 02 '14 at 17:54

phuclv · Answer 3 · 2019-06-20T02:35:58.150

The first snippet works incorrectly for negative numbers. -3.5 will be come -3, not -4. To round values properly use

int round_number(float x)
{
    if (x >= 0)
        return x + 0.5f;
    else
        return x - 0.5f
}

Even that way it's still incorrect for 2 values. See Why does Math.round(0.49999999999999994) return 1?. Note that you need to use the f suffix to get the float literal, otherwise the operation will be done in double precision and then downcast back to float

For ceiling, adding 1 is enough

int ceiling(float x)
{
    if (x < 0 || (int)x == x)
        return x;
    else
        return x + 1.0f;
}

When x is an integer, e.g. x = 3.0 (or -3.0), it returns 3 (or -3). For x = 3.1 it returns 4, for x = -3.1 it returns -3

Round up a float number that has a non-zero decimal digit

3 Answers3