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

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).

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