How to printf real digit of float in printf

Question

for instance:

#include <stdio.h>

int main(int argc, char **argv) {
    my_printf("%f\n", 1.233239208938208);
    my_printf("%f\n", 1.23);
    return (0);
};

I hope output is

1.233239208938208
1.23

How should I do?

You can print (say) 2 decimal places like this `printf("%.2f\n", 1.23);` — Weather Vane, Jan 19 '16 at 16:15
Your question is not so clear... what are you asking exactly? — Claudio Cortese, Jan 19 '16 at 16:31
I think what you want is for the computer not to print any leading zeroes, without having to manually set precision. — , Jan 19 '16 at 18:37

score 1 · Answer 1 · edited May 23 '17 at 11:52

Assuming my_printf() works like printf(), include the precision as commneted by @Weather Vane. If the precision field is not specified, 6 is used.

int main(void) {
  printf("%f\n", 1.233239208938208);
  printf("%f\n", 1.23);
  printf("%.*f\n", 15, 1.233239208938208);
  printf("%.*f\n", 2, 1.23);
  printf("%.15f\n", 1.233239208938208);
  printf("%.2f\n", 1.23);
  // Alternatively use %g which avoids trailing zeros
  printf("%.16g\n", 1.233239208938208);
  printf("%.16g\n", 1.23);
  return 0;
}

Output

1.233239
1.230000
1.233239208938208
1.23
1.233239208938208
1.23
1.233239208938208
1.23

When printing double to high precision, consider that issues begin past DBL_DIG places.
See Printf width specifier to maintain precision of floating-point value

Further detail: 1.23 is a double. Typical double, internally, do not take on the exact value as expressed in decimal text due to binary floating point representation. Instead, the closest double to 1.23 may be exactly 1.229999999999999982236431605997495353221893310546875.

How to printf real digit of float in printf

1 Answers1