How to convert a float to a string without sprintf in C?

Question

My code is taking a string format and compose it into a buffer (without sprintf, itoa etc).

I'm having trouble converting a float number to string, as I need to have precision of at most 7 digits with no trailing zeros. as well as convert each number in the float variable to char (but in that matter I don't need any help).

I tried several methods, all including math calculations, but nothing has brought me to the desired outcome. Here's my code so far, but it is messy and also sometimes gives incorrect outcome. I also believe there is a more clean and less-complicated way to do it. any help will be widely appreciated.

            if (*format == 'f') {
                float f = *(float*)ptrs[counter];
                char str[30];
                int b, c, m, k = 7, i = 0, l = 0;
                int a = f;
                f -= a;
                while (k >= 0) {
                    l = pow(10, k);
                    m = a / l;
                    if (m > 0) {
                        break;
                    }
                    k--;
                }
                printf("%d", k);
                for (l = k + 1; l > 0; l--) {
                    b = pow(10, l - 1);
                    c = a / b;
                    str[i++] = c + 48;
                    a %= b;
                }
                str[i++] = '.';
                for (l = 0; l < 7; l++) {
                    f *= 10.0;
                    b = f;
                    str[i++] = b + 48;
                    f -= b;
                }
                for (i = 0; i < 7; i++) {
                    *buffer = str[i];
                    buffer++;
                }
                counter++;
                str[i] = '\0';

for example:

input: float v2 =0.9372;
output: .937199
desired output: 0.9372

input: float v2 =0.25000;
output: 1.25000
desired output: 0.25

Show a minimal reproducible example, including the code that calls this code, the input used, the output obtained, and the output desired. — Eric Postpischil
– Eric Postpischil, Commented May 27, 2020 at 11:51
You cannot in general convert a number in a binary floating-point format to decimal using powers of ten because multiplying or dividing by powers of ten in a binary format adds rounding errors, distorting the number being converted. For a full solution, you need to use some form of extended precision or advanced techniques. If you only need to convert numbers in a limited domain, there may be easier partial solutions for that domain. You would need to specify the domain, such as numbers less than 10^6, converted with at most six significant digits. — Eric Postpischil
– Eric Postpischil, Commented May 27, 2020 at 11:57
@EricPostpischil if i understood you correctly, i need only a precision of at most 7 digits. — marksman123
– marksman123, Commented May 27, 2020 at 12:03

chux · Accepted Answer · 2020-05-27 14:44:19Z

3

he's messy and also sometimes gives incorrect outcome

At some point given the base 2 nature of typical floating point, the programmer needs to make choices:

Short code that gets close to the correct answer.
Exact correct conversion with a fair amount of code. e.g. Function to print a double - exactly
Something in between.

With common floating point encoding, there are also issues of

Infinity
Not-a-number.
Oddities like -0.0

And how portable to make the code?

Sources of inexactness

OP's use of int limits float to about [INT_MIN...INT_MAX]. Code fails for float much outside that range. Could use long long to get some more range without a great deal of code change. Better yet research float modff(float value, float *iptr).

float f = ...;
// int a=f;
long long a=f;

Repeated use of f*=10.0 below with a fractional value in f injects a possible rounding (inexactness), at each iteration.

for(l=0;l<7;l++) {
  f*=10.0;

Code makes no effort to round given that f may not be 0.0 after the fraction forming for(l=0;l<7;l++) { f*=10.0; loop. I see this as a place to improve precision. This area is also tricky as the round-up may effect many leading digits when they are 9, eventually needing a shift of the string buffer. In the 0.93721, after this loop, f was about 0.74. Since more than 0.5, a rounding would have given the answer of ".9371999" --> ".9372000".

Code aims for 7 decimal digits after the .. Values, as a decimal in code, assigned to a float match to 6 to 9 significant decimal places - which includes digits to the left of .. Expectations should not get too high.

Improvement idea

When the number of fraction digits desired is small, perform 1 scaling/round

// ASSUME f >= 0.0, (sign handling done before here)
long long a=f;
f-=a;
int int_f = round(f*pow(10, 7));
if (int_f < 10000000) {
  later_printf_the_7_digits(int_f);
} else {
  a++;
  later_printf_the_7_digits(0);
}

edited May 27, 2020 at 14:44

answered May 27, 2020 at 14:18

chux

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2 Comments

marksman123 Over a year ago

Reinstate Monice what do you exactly mean by "later_printf_the_7_digits"?

chux Over a year ago

Code here breaks f into a whole number parts and fraction part. Presumable, you want to print the whole number part first and later print the fraction part.

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