A C routine to round a float to n significant digits? - c++

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

Answer 3

Something like this should work:
double round_to_n_digits(double x, int n)
{
double scale = pow(10.0, ceil(log10(fabs(x))) + n);
return round(x * scale) / scale;
}
Alternatively you could just use sprintf/atof to convert to a string and back again:
double round_to_n_digits(double x, int n)
{
char buff[32];
sprintf(buff, "%.*g", n, x);
return atof(buff);
}
Test code for both of the above functions: http://ideone.com/oMzQZZ
Note that in some cases incorrect rounding may be observed, e.g. as pointed out by #clearScreen in the comments below, 13127.15 is rounded to 13127.1 instead of
13127.2.

Answer 4

This should work (except the noise given by floating point precision):
#include <stdio.h>
#include <math.h>
double dround(double a, int ndigits);
double dround(double a, int ndigits) {
int exp_base10 = round(log10(a));
double man_base10 = a*pow(10.0,-exp_base10);
double factor = pow(10.0,-ndigits+1);
double truncated_man_base10 = man_base10 - fmod(man_base10,factor);
double rounded_remainder = fmod(man_base10,factor)/factor;
rounded_remainder = rounded_remainder > 0.5 ? 1.0*factor : 0.0;
return (truncated_man_base10 + rounded_remainder)*pow(10.0,exp_base10) ;
}
int main() {
double a = 1.23456999;
double b = 123456.0001;
printf("%12.12f\n",dround(a,6));
printf("%12.12f\n",dround(b,6));
return 0;
}

Answer 5

If you want to print a float to a string use simple sprintf(). For outputting it just to the console you can use printf():
printf("My float is %.6f", myfloat);
This will output your float with 6 decimal places.

Answer 6

Print to 16 significant digit.
double x = -1932970.8299999994;
char buff[100];
snprintf(buff, sizeof(buff), "%.16g", x);
std::string buffAsStdStr = buff;
std::cout << std::endl << buffAsStdStr ;