Source code to round floating point number

Question

Other than using any inbuilt function like ceil and floor how can you round a floating point number?

I want to convert 2.10 to 2.14-> 2.1 and 2.15 to 2.19 -> 2.2

I wrote some logic but stuck in the middle

    float a = 2.24;
    int b = a*100;
    int result n, i =0;
    while(i>2)
    {
        n = b%10;
        i++;
    }
    if(n >5)
    {
      c = b%10;
      c = c/10 + 1;
    }

I tried lot of inbuild function but all that not workign for my diab compiler

//  totalDist = roundf(totalDist * 10) / 10.0;      
//  totalDist = floor(totalDist * pow(10., 1) + .5) / pow(10., 1);  
    totalDist = floorf(totalDist * 10 + 0.5) / 10;

planning to write my own logic

Answer 1

float a = 2.24;
float b = roundf(a*10.0f)/10.0f;//roundf in C99

Answer 2

Try int((x + 0.05) * 10.0) / 10.0 .

You can also use trunc instead of the conversion to and from int .

Or you could use round : round(x * 10.0) / 10.0

Answer 3

If you just want to round the number for output purposes, then the %.1f format string is indeed the correct answer.

printf("%.1f", 2.14);

or If you actually want to round off value then something like this work

#include <math.h>

float val = 37.777779;

float rounded_down = floorf(val * 100) / 100;   /* Result: 37.77 */
float nearest = floorf(val * 100 + 0.5) / 100;  /* Result: 37.78 */
float rounded_up = ceilf(val * 100) / 100;      /* Result: 37.78 */

Usually we use nearest method

Answer 4

BLUEPIXY's answer is good. But in general; floating point multiplication is faster than division, so if speed is relevant; multiply by 0.1 instead of dividing by 10.

Answer 5

Because my old answer needs lot of careful programming to make it work

I present another way to implement an round function:

You can easily check if a (positive/negative) number will be rounded up or down by (adding/subtracting) 0.5. Then you can truncate the result and take your answer. Also, in order to support deeper level of round as you want you can let the user specify how many decimal places of round the want, and voila:

long double round_f(long double number, int decimal_places)
{
     assert(decimal_places > 0);

     double power = pow(10, decimal_places-1);
     number *= power;

     return (number >= 0) ? ((long long)(number + 0.5))/power : ((long long)(number - 0.5))/power;
}

And here you can see a working example.

Old one:

You can use sprint:

#include <stdio.h>

int main(void)
{
    float a = 2.24;
    int size;
    char num[8] = {0};
    sprintf(num, "%d", (int) (a * 100));

    for (size=0; num[size]; size++);

    size--;

    if (num[size] < '5') {
        num[size] = '0';
    } else {
        num[size] = '0';
        if (num[size-1] == '9')
            num[size-1] = '0';
        else
            num[size-1]++;
    }

    printf("a = %f\n", a);
    printf("num = %s\n", num);

    return(0);
}

You can see the logic. Finally you can revert the string to integer and multiply by 100.

EDIT: It's only the logic. See comments below for what is needed to fully work.