Quick C question - how to hide leading zero in printf

Question

Hey, if I had this code:

printf( "%8.2f" , .23 );

It outputs 0.23. How do I get it to simply output .23?

Thanks.

Answer 1

The C standard says that for the f and F floating point format specifiers:

If a decimal-point character appears, at least one digit appears before it.

I think that if you don't want a zero to appear before the decimal point, you'll probably have to do something like use snprintf() to format the number into a string, and remove the 0 if the formatted string starts with "0." (and similarly for "-0."). Then pass that formatted string to our real output. Or something like that.

Answer 2

it is not possible to do it only using printf. The printf standard says :

f  - "double" argument is output in conventional form, i.e.
     [-]mmmm.nnnnnn
     The default number of digits after the decimal point is six,
     but this can be changed with a precision field. If a decimal point
     appears, at least one digit appears before it. The "double" value is
     rounded to the correct number of decimal places.

Note the If a decimal point appears, at least one digit appears before it

Therefore it seems you have to handcode your own formatter.

Answer 3

This'll do the trick:

double a;
printf( ".%d", (int)modf(0.23,&a) );

Answer 4

double f = 0.23;

assert(f < 1 && f >= 0);  
printf(".%u\n" , (unsigned)((f + 0.005) * 100));

Answer 5

There isn't any format string specifier to remove that leading 0, so you'll have to roll your own; This is the shortest thing I can think of:

char s[10];
sprintf(s, "%.2f", .23);
printf("%8s", (s[0] == '0' ? &s[1] : s));

Answer 6

#include <stdio.h>

static void printNoLeadingZeros(double theValue)
{
   char buffer[255] = { '\0' };

   sprintf(buffer, "%.2f", theValue);

   printf("%s\n", buffer + (buffer[0] == '0'));
}

int main()
{
   double values[] = { 0.23, .23, 1.23, 01.23, 001.23, 101.23 };
   int n           = sizeof(values) / sizeof(values[0]);
   int i           = 0;

   while(i < n)
      printNoLeadingZeros(values[i++]);

   return(0);
}

Answer 7

It looks there is no easy solution. I would probably use something like code below. It is not the fastest method, however it should work with many different formats. It preserves number of char and position of dot too.

#include <stdio.h>

void fixprint(char *s)
{
        size_t i;
        i = 1;
        while (s[i]=='0' || s[i]==' ' || s[i]=='+' || s[i]=='-') {
                if (s[i]=='0') s[i]=' ';
                i++;
        }
}

int main()
{
        float x = .23;
        char s[14];
        sprintf(s,"% 8.2f",x);
        fixprint(s);
        printf("%s\n",s);
}