A C routine to round a float to n significant digits?

Question

Suppose I have a float. I would like to round it to a certain number of significant digits.

In my case n=6.

So say float was f=1.23456999;

round(f,6) would give 1.23457

f=123456.0001 would give 123456

Anybody know such a routine ?

Here it works on website: http://ostermiller.org/calc/significant_figures.html

Answer 1

Multiply the number by a suitable scaling factor to move all significant digits to the left of the decimal point. Then round and finally reverse the operation:

#include <math.h>

double round_to_digits(double value, int digits)
{
    if (value == 0.0) // otherwise it will return 'nan' due to the log10() of zero
        return 0.0;

    double factor = pow(10.0, digits - ceil(log10(fabs(value))));
    return round(value * factor) / factor;   
}

Tested: http://ideone.com/fH5ebt

Buts as @PascalCuoq pointed out: the rounded value may not exactly representable as a floating point value.

Answer 2

#include <stdio.h> 
#include <string.h>
#include <stdlib.h>

char *Round(float f, int d)
{
    char buf[16];
    sprintf(buf, "%.*g", d, f);
    return strdup(buf);
}

int main(void)
{
    char *r = Round(1.23456999, 6);
    printf("%s\n", r);
    free(r);
}

Output is:

1.23457