Task: Long multiplication





Java: public class LongMult {

	private static byte[] stringToDigits(String num) {
		byte[] result = new byte[num.length()];
		for (int i = 0; i < num.length(); i++) {
			char c = num.charAt(i);
			if (c < '0' || c > '9') {
				throw new IllegalArgumentException("Invalid digit " + c
						+ " found at position " + i);
			}
			result[num.length() - 1 - i] = (byte) (c - '0');
		}
		return result;
	}

	public static String longMult(String num1, String num2) {
		byte[] left = stringToDigits(num1);
		byte[] right = stringToDigits(num2);
		byte[] result = new byte[left.length + right.length];
		for (int rightPos = 0; rightPos < right.length; rightPos++) {
			byte rightDigit = right[rightPos];
			byte temp = 0;
			for (int leftPos = 0; leftPos < left.length; leftPos++) {
				temp += result[leftPos + rightPos];
				temp += rightDigit * left[leftPos];
				result[leftPos + rightPos] = (byte) (temp % 10);
				temp /= 10;
			}
			int destPos = rightPos + left.length;
			while (temp != 0) {
				temp += result[destPos] & 0xFFFFFFFFL;
				result[destPos] = (byte) (temp % 10);
				temp /= 10;
				destPos++;
			}
		}
		StringBuilder stringResultBuilder = new StringBuilder(result.length);
		for (int i = result.length - 1; i >= 0; i--) {
			byte digit = result[i];
			if (digit != 0 || stringResultBuilder.length() > 0) {
				stringResultBuilder.append((char) (digit + '0'));
			}
		}
		return stringResultBuilder.toString();
	}

	public static void main(String[] args) {
		System.out.println(longMult("18446744073709551616",
				"18446744073709551616"));
	}
}




VB: Imports System
Imports System.Console
Imports BI = System.Numerics.BigInteger

Module Module1

    Dim a As Decimal, mx As Decimal = 1E28D, hm As Decimal = 1E14D

    
    Structure bd
        Public hi, lo As Decimal
    End Structure

    
    Function toStr(ByVal a As bd, ByVal Optional comma As Boolean = False) As String
        Dim r As String = If(a.hi = 0, String.Format("{0:0}", a.lo),
                                       String.Format("{0:0}{1:" & New String("0"c, 28) & "}", a.hi, a.lo))
        If Not comma Then Return r 
        Dim rc As String = ""
        For i As Integer = r.Length - 3 To 0 Step -3
            rc = "," & r.Substring(i, 3) & rc : Next
        toStr = r.Substring(0, r.Length Mod 3) & rc
            toStr = toStr.Substring(If(toStr.Chars(0) = "," , 1, 0))
    End Function

    
    Function Pow_dec(ByVal bas As Decimal, ByVal exp As UInteger) As Decimal
        If exp = 0 Then Pow_dec = 1D else Pow_dec = Pow_dec(bas, exp >> 1) : _
        Pow_dec *= Pow_dec : If (exp And 1) <> 0 Then Pow_dec *= bas
    End Function

    Sub Main(ByVal args As String())
         For p As UInteger = 64 To 95 - 1 Step 30                 
            Dim y As bd, x As bd : a = Pow_dec(2D, p)             
            WriteLine("The square of (2^{0}):                    {1,38:n0}", p, a)
            x.hi = Math.Floor(a / hm) : x.lo = a Mod hm           
            Dim BS As BI = BI.Pow(CType(a, BI), 2)                
            y.lo = x.lo * x.lo : y.hi = x.hi * x.hi               
            a = x.hi * x.lo * 2D                                  
            y.hi += Math.Floor(a / hm) : y.lo += (a Mod hm) * hm  
            While y.lo > mx : y.lo -= mx : y.hi += 1 : End While  
            WriteLine(" is {0,75} (which {1} match the BigInteger computation)" & vbLf,
                toStr(y, True), If(BS.ToString() = toStr(y), "does", "fails to"))
        Next
    End Sub

End Module