Get Integer and Fractional part of a decimal number

Question

Given X.Y, I want to get X and Y.

For instance, given 123.456 I want to get 123 and 456 (NOT 0.456).

I can do the following:

decimal num = 123.456M;
decimal integer = Math.Truncate(num);
decimal fractional = num - Math.Truncate(num);    

// integer = 123
// fractional = 0.456   but I want 456

REF

As above-mentioned, using this method I will get 0.456, while I need 456. Sure I can do the following:

int fractionalNums = (int)((num - Math.Truncate(num)) * 1000); 

// fracionalNums = 456

Additionally, this method requires knowing how many fractional numbers a given decimal number has so that you can multiply to that number (e.g., 123.456 has three, 123.4567 has four, 123.456789 has six, 123.1234567890123456789 has nineteen).

Few points to consider:

• This operation will be executed millions of times; hence, performance is critical (maybe a bit-wise-based solution would do better);
• Precision is critical, and no rounding is acceptable.

---

NOTE 1

For performance reasons, I am NOT interested in string manipulation-based approaches.

---

NOTE 2

The numbers in my question are of decimal type, hence methods that work for only decimal types and fail on float or double (due to floating point precision) are acceptable.

---

NOTE 3

Two sides of decimal (i.e., integer and fractional parts) can be considered two integers. Hence, 123.000456 is not an expected input; and even if it is given, it is acceptable to split it to 123 and 456 (because both sides are to be considered integers).

Answer 1

BitConverter.GetBytes(decimal.GetBits(num)[3])[2]; - number of digits after comma

long[] tens = new long[] {1, 10, 100, 1000, ...};

decimal num = 123.456M;
int iPart = (int)num;
decimal dPart = num - iPart;
int count = BitConverter.GetBytes(decimal.GetBits(num)[3])[2];

long pow = tens[count];

Console.WriteLine(iPart);
Console.WriteLine((long)(dPart * pow));

Answer 2

Decimal has a 96 bit mantissa, so a long is not good enough to get every possible value.

Define all (positive) powers of 10 defined for Decimal:

decimal mults[] = {1M, 1e1M, 1e2M, 1e3M, <insert rest here>, 1e27M, 1e28M};

Then, inside the loop you need to get the scale (the power of 10 by which the "mantissa" is divided to get the nominal value of the decimal):

int[] bits = Decimal.GetBits(n);
int scale = (bits[3] >> 16) & 31;               // 567.1234 represented as 5671234 x 10^-4

decimal intPart = (int)n;                       // 567.1234 --> 567
decimal decPart = (n - intPart) * mults[scale]; // 567.1234 --> 0.1234 --> 1234

Answer 3

The easiest way is probably to convert the number to string. Then take the substring after the decimal point, and convert it back to int.