Convert double/float to string

Solution 1:

Use snprintf() from stdlib.h. Worked for me.

double num = 123412341234.123456789; 
char output[50];

snprintf(output, 50, "%f", num);

printf("%s", output);

Solution 2:

The only exact solution is to perform arbitrary-precision decimal arithmetic for the base conversion, since the exact value can be very long - for 80-bit long double, up to about 10000 decimal places. Fortunately it's "only" up to about 700 places or so for IEEE double.

Rather than working with individual decimal digits, it's helpful to instead work base-1-billion (the highest power of 10 that fits in a 32-bit integer) and then convert these "base-1-billion digits" to 9 decimal digits each at the end of your computation.

I have a very dense (rather hard to read) but efficient implementation here, under LGPL MIT license:

http://git.musl-libc.org/cgit/musl/blob/src/stdio/vfprintf.c?h=v1.1.6

If you strip out all the hex float support, infinity/nan support, %g/%f/%e variation support, rounding (which will never be needed if you only want exact answers), and other things you might not need, the remaining code is rather simple.

Solution 3:

Go and look at the printf() implementation with "%f" in some C library.

Solution 4:

I know maybe it is unnecessary, but I made a function which converts float to string:

CODE:

#include <stdio.h>

/** Number on countu **/

int n_tu(int number, int count)
{
    int result = 1;
    while(count-- > 0)
        result *= number;

    return result;
}

/*** Convert float to string ***/
void float_to_string(float f, char r[])
{
    long long int length, length2, i, number, position, sign;
    float number2;

    sign = -1;   // -1 == positive number
    if (f < 0)
    {
        sign = '-';
        f *= -1;
    }

    number2 = f;
    number = f;
    length = 0;  // Size of decimal part
    length2 = 0; // Size of tenth

    /* Calculate length2 tenth part */
    while( (number2 - (float)number) != 0.0 && !((number2 - (float)number) < 0.0) )
    {
         number2 = f * (n_tu(10.0, length2 + 1));
         number = number2;

         length2++;
    }

    /* Calculate length decimal part */
    for (length = (f > 1) ? 0 : 1; f > 1; length++)
        f /= 10;

    position = length;
    length = length + 1 + length2;
    number = number2;
    if (sign == '-')
    {
        length++;
        position++;
    }

    for (i = length; i >= 0 ; i--)
    {
        if (i == (length))
            r[i] = '\0';
        else if(i == (position))
            r[i] = '.';
        else if(sign == '-' && i == 0)
            r[i] = '-';
        else
        {
            r[i] = (number % 10) + '0';
            number /=10;
        }
    }
}

Solution 5:

See if the BSD C Standard Library has fcvt(). You could start with the source for it that rather than writing your code from scratch. The UNIX 98 standard fcvt() apparently does not output scientific notation so you would have to implement it yourself, but I don't think it would be hard.