Use printf to format floats without decimal places if only trailing 0s

Question

Is it possible to format a float in C to only show up to 2 decimal places if different from 0s using printf?

Ex:

12 => 12

12.1 => 12.1

12.12 => 12.12

I tried using:

float f = 12; printf("%.2f", f) 

but I get

12 => 12.00

12.1 => 12.10

12.12 => 12.12

Answer 1

You can use the %g format specifier:

#include <stdio.h>  int main() {   float f1 = 12;   float f2 = 12.1;   float f3 = 12.12;   float f4 = 12.1234;   printf("%g\n", f1);   printf("%g\n", f2);   printf("%g\n", f3);   printf("%g\n", f4);   return 0; } 

Result:

 12 12.1 12.12 12.1234 

Note that, unlike the f format specifier, if you specify a number before the g it refers to the length of the entire number (not the number of decimal places as with f).

Answer 2

From our discussion in the above answer here is my program that works for any number of digits before the decimal.

#include <stdio.h> #include <stdlib.h> #include <string.h>  int main() {     float f1 = 12.13;     float f2 = 12.245;     float f3 = 1242.145;     float f4 = 1214.1;      int i = 0;     char *s1 = (char *)(malloc(sizeof(char) * 20));     char *s2 = (char *)(malloc(sizeof(char) * 20));      sprintf(s1, "%f", f1);     s2 = strchr(s1, '.');     i = s2 - s1;     printf("%.*g\n", (i+2), f1);      sprintf(s1, "%f", f2);     s2 = strchr(s1, '.');     i = s2 - s1;     printf("%.*g\n", (i+2), f2);      sprintf(s1, "%f", f3);     s2 = strchr(s1, '.');     i = s2 - s1;     printf("%.*g\n", (i+2), f3);      sprintf(s1, "%f", f4);     s2 = strchr(s1, '.');     i = s2 - s1;     printf("%.*g\n", (i+2), f4);      free(s1);     free(s2);      return 0; } 

And here's the output

12.13 12.24 1242.15 1214.1