Wrong logarithm of BigInteger when size of BigInteger exceeds ¼ gigabyte

Question

When I have a BigInteger whose size exceeds 2 gigabits (that's ¼ gigabyte; I found this threshold by trial and error), the logarithm method gives a wrong answer. This simple code illustrates:

  byte[] bb;

  bb = new byte[150000001];
  bb[150000000] = 1;  // sets most significant byte to one
  var i1 = new BigInteger(bb);
  double log1 = BigInteger.Log(i1);
  Console.WriteLine(log1);   // OK, writes 831776616.671934

  bb = new byte[300000001];
  bb[300000000] = 1;  // sets most significant byte to one
  var i2 = new BigInteger(bb);
  double log2 = BigInteger.Log(i2);
  Console.WriteLine(log2);   // Broken, gives negative number, should be twice 831776616.671934

Of course we must have a positive log for a number exceeding 1, a zero log for the number 1, and a negative log for a number between 0 and 1 (no integers there). My numbers i1 and i2 above are greater than 1 since, by convention, when the most significant byte is between 0 and 127, that means positive BigInteger.

Now, if you read the documentation for BigInteger.Log, they claim it might throw if the logarithm "is out of range of the Double data type". Now, clearly that would require a computer with a memory storage of more than 1E+300 bytes, and the observable universe is much too small to contain such a computer, so I guess that will never happen.

So why doesn't this work?

PS! Size over 2 ^^ 31 bits means that the actual value of the BigInteger is over 2 ^^ (2 ^^ 31), or approximately circa 8.8E+646456992.

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UPDATE: I sent in a bug report to Microsoft Connect. After having read the discussions I have also become aware that because of the design of BigInteger and the upper limit of 2 gigabyte for the size of one single object, a BigInteger can never be over 2 gigabyte (no matter how much memory you have). This bug happens, therefore, when the BigInteher is between ¼ and 2 gigabytes.

Answer 1

Let me guess: The value is

-1.3134912384757032e9

(modulo small variations in computing the logarithm)?

The index of the highest set bit is stored and passed in an int, and

8*300000000 = 2400000000 > 2147483647

so the index wraps around to a negative number, namely -1894967296, and

-1894967296 * log 2 = -1.3134912384757032e9

Oops. Somebody should file a bug report.