Procedure C for rounding 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, the float was f=1.23456999;

round(f,6) will give 1.23457

f=123456.0001 will give 123456

Does anyone know such a routine?

Here he works on the 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 cancel 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

But, as @PascalCuoq pointed out, the rounded value may not display exactly 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); } 

Exit:

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 can simply use sprintf / atof to convert to string and vice versa:

 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

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Please note that in some cases incorrect rounding may occur, for example. as pointed out by @clearScreen in the comments below, 13127.15 is rounded to 13127.1 instead of 13127.2.

Answer 4

If you want to print a float in a string, use simple sprintf() . To output it to the console, you can use printf() :

 printf("My float is %.6f", myfloat); 

This will display your float with six decimal places.

Answer 5

This should work (except for the noise set using 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 6

Print up to 16 significant digits.

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