Get a numer decimal part as Integer using only math

Question

Edit: This has to do with how computers handle floating point operations, a fact that every programmer faces once in a lifetime. I didn't understand this correctly when I asked the question.

I know the simplest way to start dealing with this would be:

val floatNumber: Float = 123.456f
val decimalPart = floatNumber - floatNumber.toInt() //This would be 0.456 (I don't care about precision as this is not the main objective of my question)

Now in a real world with a pen and a piece of paper, if I want to "convert" the decimal part 0.456 to integer, I just need to multiply 0.456 * 1000, and I get the desired result, which is 456 (an integer number).

Many proposed solutions suggest splitting the number as string and extracting the decimal part this way, but I need the solution to be obtained mathematically, not using strings.

Given a number, with an unknown number of decimals (convert to string and counting chars after . or , is not acceptable), I need to "extract" it's decimal part as an integer using only math.

Read questions like this with no luck:

How to get the decimal part of a float?

How to extract fractional digits of double/BigDecimal

If someone knows a kotlin language solution, it would be great. I will post this question also on the math platform just in case. How do I get whole and fractional parts from double in JSP/Java?

Update: Is there a "mathematical" way to "calculate" how many decimals a number has? (It is obvious when you convert to string and count the chars, but I need to avoid using strings) It would be great cause calculating: decimal (0.456) * 10 * number of decimals(3) will produce the desired result.

Update 2 This is not my use-case, but I guess it will clarify the idea: Suppose you want to calculate a constant(such as PI), and want to return an integer with at most 50 digits of the decimal part of the constant. The constant doesn't have to be necessarily infinite (can be for example 0.5, in which case "5" will be returned)

Answer 1

I would just multiply the fractional number by 10 (or move the decimal point to the right) until it has no fractional part left:

public static long fractionalDigitsLong(BigDecimal value) {
    BigDecimal fractional = value.remainder(BigDecimal.ONE);
    long digits;
    do {
        fractional = fractional.movePointRight(1);  // or multiply(BigDecimal.TEN)
        digits = fractional.longValue();
    } while (fractional.compareTo(BigDecimal.valueOf(digits)) != 0);
    return digits;
}

Note 1: using BigDecimal to avoid floating point precision problems

Note 2: using compareTo since equals also compares the scale ("0.0" not equals "0.00")

^{(sure the BigDecimal already knows the size of the fractional part, just the value returned by scale())}

---

Complement:

If using BigDecimal the whole problem can be compressed to:

public static BigInteger fractionalDigits(BigDecimal value) {
    return value.remainder(BigDecimal.ONE).stripTrailingZeros().unscaledValue();
}

^{stripping zeros can be suppressed if desired}

Answer 2

I am not sure if it counts against you on this specific problem if you use some String converters with a method(). That is one way to get the proper answer. I know that you stated you couldn't use String, but would you be able to use Strings within a Custom made method? That could get you the answer that you need with precision. Here is the class that could help us convert the number:

class NumConvert{
     String theNum;

     public NumConvert(String theNum) {
           this.theNum = theNum;
}
     public int convert() {
            String a = String.valueOf(theNum);
            String[] b = a.split("\\.");
            String b2 = b[1];
            int zeros = b2.length();
            String num = "1";
            for(int x = 0; x < zeros; x++) {
                num += "0";
    }
            float c = Float.parseFloat(theNum);
            int multiply = Integer.parseInt(num);
            float answer = c - (int)c;
            int integerForm = (int)(answer * multiply);
            return integerForm;
     }
 }

Then within your main class:

public class ChapterOneBasics {
    public static void main(String[] args) throws java.io.IOException{
          NumConvert n = new NumConvert("123.456");
          NumConvert q = new NumConvert("123.45600128");
          System.out.println(q.convert());
          System.out.println(n.convert());
    }
}

output:

 45600128
 456

Answer 3

Float or Double are imprecise, just an approximation - without precision. Hence 12.345 is somewhere between 12.3449... and 12.3450... .

This means that 12.340 cannot be distinghuished from 12.34. The "decimal part" would be 34 divided by 100. Also 12.01 would have a "decimal part" 1 divided by 100, and too 12.1 would have 1 divided by 10.

So a complete algorith would be (using java):

int[] decimalsAndDivider(double x) {
    int decimalPart = 0;
    int divider = 1;
    final double EPS = 0.001;
    for (;;) {
        double error = x - (int)x;
        if (-EPS < error && error < EPS) {
            break;
        }
        x *= 10;
        decimalPart = 10 * decimalPart + ((int)(x + EPS) % 10);
        divider *= 10;
    }
    return new int[] { decimalPart, divider };
}

Answer 4

I posted the below solution yesterday after testing it for a while, and later found that it does not always work due to problems regarding precision of floats, doubles and bigdecimals.
My conclusion is that this problem is unsolvable if you want infinite precision:
So I re-post the code just for reference:

fun getDecimalCounter(d: Double): Int {
    var temp = d
    var tempInt = Math.floor(d)

    var counter = 0

    while ((temp - tempInt)  > 0.0 ) {
        temp *= 10
        tempInt = Math.floor(temp)
        counter++
    }

    return counter
}


fun main(args: Array <String> ) {
    var d = 3.14159
    if (d < 0) d = -d
    val decimalCounter = getDecimalCounter(d)
    val decimalPart = (d - Math.floor(d))
    var decimalPartInt = Math.round(decimalPart * 10.0.pow(decimalCounter))
    while (decimalPartInt % 10 == 0L) {
        decimalPartInt /= 10
    }
    println(decimalPartInt)
}

I dropped floats because of lesser precision and used doubles.
The final rounding is also necessary due to precision.