Convert float to string without sprintf()

Question

I'm coding for a microcontroller-based application and I need to convert a float to a character string, but I do not need the heavy overhead associated with sprintf(). Is there any eloquent way to do this? I don't need too much. I only need 2 digits of precision.

Answer 1

Here's a version optimized for embedded systems that doesn't require any stdio or memset, and has low memory footprint. You're responsible for passing a char buffer initialized with zeros (with pointer p) where you want to store your string, and defining CHAR_BUFF_SIZE when you make said buffer (so the returned string will be null terminated).

static char * _float_to_char(float x, char *p) {
    char *s = p + CHAR_BUFF_SIZE; // go to end of buffer
    uint16_t decimals;  // variable to store the decimals
    int units;  // variable to store the units (part to left of decimal place)
    if (x < 0) { // take care of negative numbers
        decimals = (int)(x * -100) % 100; // make 1000 for 3 decimals etc.
        units = (int)(-1 * x);
    } else { // positive numbers
        decimals = (int)(x * 100) % 100;
        units = (int)x;
    }

    *--s = (decimals % 10) + '0';
    decimals /= 10; // repeat for as many decimal places as you need
    *--s = (decimals % 10) + '0';
    *--s = '.';

    while (units > 0) {
        *--s = (units % 10) + '0';
        units /= 10;
    }
    if (x < 0) *--s = '-'; // unary minus sign for negative numbers
    return s;
}

Tested on ARM Cortex M0 & M4. Rounds correctly.

Answer 2

Try this. It should be nice and small. I've output the string directly - doing a printf, rather than a sprintf. I'll leave it to you to allocate space for the return string, as well as copying the result into it.

// prints a number with 2 digits following the decimal place
// creates the string backwards, before printing it character-by-character from
// the end to the start
//
// Usage: myPrintf(270.458)
//  Output: 270.45
void myPrintf(float fVal)
{
    char result[100];
    int dVal, dec, i;

    fVal += 0.005;   // added after a comment from Matt McNabb, see below.

    dVal = fVal;
    dec = (int)(fVal * 100) % 100;

    memset(result, 0, 100);
    result[0] = (dec % 10) + '0';
    result[1] = (dec / 10) + '0';
    result[2] = '.';

    i = 3;
    while (dVal > 0)
    {
        result[i] = (dVal % 10) + '0';
        dVal /= 10;
        i++;
    }

    for (i=strlen(result)-1; i>=0; i--)
        putc(result[i], stdout);
}