如何将数字转换为BouncyCastle.BigInteger＆lt; - ＆gt; System.Numerics.BigInteger

Question

我的代码中最慢的部分是我使用Bouncy Castle的“BigInteger.Multiply（）”方法。为了尝试加快速度，我试图插入下面的代码，看看是否有任何改进。

我知道我做的不对，因为当Bouncy Castle看到我发送给它的BigInteger时，我会遇到很多错误和异常：

         if (this.sign != 1 || val.sign != 1)
        {
            //todo: test encoding, twos compliment
        }
        var tempThis = this.ToByteArrayUnsigned();
        Array.Reverse(tempThis);
        System.Numerics.BigInteger aThis = new System.Numerics.BigInteger(tempThis);

        var tempThat = val.ToByteArrayUnsigned();
        Array.Reverse(tempThat);
        System.Numerics.BigInteger bThat = new System.Numerics.BigInteger(tempThat);
        var msResult = System.Numerics.BigInteger.Multiply(aThis, bThat);

        var tmpRet = msResult.ToByteArray();
        Array.Reverse(tmpRet);
        var ret = new BigInteger(tmpRet);
        return ret;

这是原始BC功能

    public BigInteger Multiply(     BigInteger val)
    {

       // THEORY: If I use the native System.Numerics here, then I might get better performance

        if (val == this)
            return Square();

        if ((sign & val.sign) == 0)
            return Zero;

        if (val.QuickPow2Check()) // val is power of two
        {
            BigInteger result = this.ShiftLeft(val.Abs().BitLength - 1);
            return val.sign > 0 ? result : result.Negate();
        }

        if (this.QuickPow2Check()) // this is power of two
        {
            BigInteger result = val.ShiftLeft(this.Abs().BitLength - 1);
            return this.sign > 0 ? result : result.Negate();
        }

        int resLength = magnitude.Length + val.magnitude.Length;
        int[] res = new int[resLength];

        Multiply(res, this.magnitude, val.magnitude);

        int resSign = sign ^ val.sign ^ 1;
        return new BigInteger(resSign, res, true);
    }