bigunsigned.cc

实现了大整数的四则运算的基础类,BIgInt类的设计与实现

	* For each 1-bit of `a' (say the `i2'th bit of block `i'):
	*    Add `b << (i blocks and i2 bits)' to *this.
	*/
	// Variables for the calculation
	Index i, j, k;
	unsigned int i2;
	Blk temp;
	bool carryIn, carryOut;
	// Set preliminary length and make room
	len = a.len + b.len;
	allocate(len);
	// Zero out this object
	for (i = 0; i < len; i++)
		blk[i] = 0;
	// For each block of the first number...
	for (i = 0; i < a.len; i++) {
		// For each 1-bit of that block...
		for (i2 = 0; i2 < N; i2++) {
			if ((a.blk[i] & (Blk(1) << i2)) == 0)
				continue;
			/*
			* Add b to this, shifted left i blocks and i2 bits.
			* j is the index in b, and k = i + j is the index in this.
			*
			* `getShiftedBlock', a short inline function defined above,
			* is now used for the bit handling.  It replaces the more
			* complex `bHigh' code, in which each run of the loop dealt
			* immediately with the low bits and saved the high bits to
			* be picked up next time.  The last run of the loop used to
			* leave leftover high bits, which were handled separately.
			* Instead, this loop runs an additional time with j == b.len.
			* These changes were made on 2005.01.11.
			*/
			for (j = 0, k = i, carryIn = false; j <= b.len; j++, k++) {
				/*
				* The body of this loop is very similar to the body of the first loop
				* in `add', except that this loop does a `+=' instead of a `+'.
				*/
				temp = blk[k] + getShiftedBlock(b, j, i2);
				carryOut = (temp < blk[k]);
				if (carryIn) {
					temp++;
					carryOut |= (temp == 0);
				}
				blk[k] = temp;
				carryIn = carryOut;
			}
			// No more extra iteration to deal with `bHigh'.
			// Roll-over a carry as necessary.
			for (; carryIn; k++) {
				blk[k]++;
				carryIn = (blk[k] == 0);
			}
		}
	}
	// Zap possible leading zero
	if (blk[len - 1] == 0)
		len--;
}

/*
* DIVISION WITH REMAINDER
* The functionality of divide, modulo, and %= is included in this one monstrous call,
* which deserves some explanation.
*
* The division *this / b is performed.
* Afterwards, q has the quotient, and *this has the remainder.
* Thus, a call is like q = *this / b, *this %= b.
*
* This seemingly bizarre pattern of inputs and outputs has a justification.  The
* ``put-here operations'' are supposed to be fast.  Therefore, they accept inputs
* and provide outputs in the most convenient places so that no value ever needs
* to be copied in its entirety.  That way, the client can perform exactly the
* copying it needs depending on where the inputs are and where it wants the output.
* A better name for this function might be "modWithQuotient", but I would rather
* not change the name now.
*/
void BigUnsigned::divideWithRemainder(const BigUnsigned &b, BigUnsigned &q) {
	/*
	 * Defending against aliased calls is a bit tricky because we are
	 * writing to both *this and q.
	 * 
	 * It would be silly to try to write quotient and remainder to the
	 * same variable.  Rule that out right away.
	 */
	if (this == &q)
		throw "BigUnsigned::divideWithRemainder: Cannot write quotient and remainder into the same variable";
	/*
	 * Now *this and q are separate, so the only concern is that b might be
	 * aliased to one of them.  If so, use a temporary copy of b.
	 */
	if (this == &b || &q == &b) {
		BigUnsigned tmpB(b);
		divideWithRemainder(tmpB, q);
		return;
	}
	
	/*
	* Note that the mathematical definition of mod (I'm trusting Knuth) is somewhat
	* different from the way the normal C++ % operator behaves in the case of division by 0.
	* This function does it Knuth's way.
	*
	* We let a / 0 == 0 (it doesn't matter) and a % 0 == a, no exceptions thrown.
	* This allows us to preserve both Knuth's demand that a mod 0 == a
	* and the useful property that (a / b) * b + (a % b) == a.
	*/
	if (b.len == 0) {
		q.len = 0;
		return;
	}
	
	/*
	* If *this.len < b.len, then *this < b, and we can be sure that b doesn't go into
	* *this at all.  The quotient is 0 and *this is already the remainder (so leave it alone).
	*/
	if (len < b.len) {
		q.len = 0;
		return;
	}
	
	/*
	* At this point we know *this > b > 0.  (Whew!)
	*/
	
	/*
	* Overall method:
	*
	* For each appropriate i and i2, decreasing:
	*    Try to subtract (b << (i blocks and i2 bits)) from *this.
	*        (`work2' holds the result of this subtraction.)
	*    If the result is nonnegative:
	*        Turn on bit i2 of block i of the quotient q.
	*        Save the result of the subtraction back into *this.
	*    Otherwise:
	*        Bit i2 of block i remains off, and *this is unchanged.
	* 
	* Eventually q will contain the entire quotient, and *this will
	* be left with the remainder.
	*
	* We use work2 to temporarily store the result of a subtraction.
	* work2[x] corresponds to blk[x], not blk[x+i], since 2005.01.11.
	* If the subtraction is successful, we copy work2 back to blk.
	* (There's no `work1'.  In a previous version, when division was
	* coded for a read-only dividend, `work1' played the role of
	* the here-modifiable `*this' and got the remainder.)
	*
	* We never touch the i lowest blocks of either blk or work2 because
	* they are unaffected by the subtraction: we are subtracting
	* (b << (i blocks and i2 bits)), which ends in at least `i' zero blocks.
	*/
	// Variables for the calculation
	Index i, j, k;
	unsigned int i2;
	Blk temp;
	bool borrowIn, borrowOut;
	
	/*
	* Make sure we have an extra zero block just past the value.
	*
	* When we attempt a subtraction, we might shift `b' so
	* its first block begins a few bits left of the dividend,
	* and then we'll try to compare these extra bits with
	* a nonexistent block to the left of the dividend.  The
	* extra zero block ensures sensible behavior; we need
	* an extra block in `work2' for exactly the same reason.
	*
	* See below `divideWithRemainder' for the interesting and
	* amusing story of this section of code.
	*/
	Index origLen = len; // Save real length.
	// 2006.05.03: Copy the number and then change the length!
	allocateAndCopy(len + 1); // Get the space.
	len++; // Increase the length.
	blk[origLen] = 0; // Zero the extra block.
	
	// work2 holds part of the result of a subtraction; see above.
	Blk *work2 = new Blk[len];
	
	// Set preliminary length for quotient and make room
	q.len = origLen - b.len + 1;
	q.allocate(q.len);
	// Zero out the quotient
	for (i = 0; i < q.len; i++)
		q.blk[i] = 0;
	
	// For each possible left-shift of b in blocks...
	i = q.len;
	while (i > 0) {
		i--;
		// For each possible left-shift of b in bits...
		// (Remember, N is the number of bits in a Blk.)
		q.blk[i] = 0;
		i2 = N;
		while (i2 > 0) {
			i2--;
			/*
			* Subtract b, shifted left i blocks and i2 bits, from *this,
			* and store the answer in work2.  In the for loop, `k == i + j'.
			*
			* Compare this to the middle section of `multiply'.  They
			* are in many ways analogous.  See especially the discussion
			* of `getShiftedBlock'.
			*/
			for (j = 0, k = i, borrowIn = false; j <= b.len; j++, k++) {
				temp = blk[k] - getShiftedBlock(b, j, i2);
				borrowOut = (temp > blk[k]);
				if (borrowIn) {
					borrowOut |= (temp == 0);
					temp--;
				}
				// Since 2005.01.11, indices of `work2' directly match those of `blk', so use `k'.
				work2[k] = temp; 
				borrowIn = borrowOut;
			}
			// No more extra iteration to deal with `bHigh'.
			// Roll-over a borrow as necessary.
			for (; k < origLen && borrowIn; k++) {
				borrowIn = (blk[k] == 0);
				work2[k] = blk[k] - 1;
			}
			/*
			* If the subtraction was performed successfully (!borrowIn),
			* set bit i2 in block i of the quotient.
			*
			* Then, copy the portion of work2 filled by the subtraction
			* back to *this.  This portion starts with block i and ends--
			* where?  Not necessarily at block `i + b.len'!  Well, we
			* increased k every time we saved a block into work2, so
			* the region of work2 we copy is just [i, k).
			*/
			if (!borrowIn) {
				q.blk[i] |= (Blk(1) << i2);
				while (k > i) {
					k--;
					blk[k] = work2[k];
				}
			} 
		}
	}
	// Zap possible leading zero in quotient
	if (q.blk[q.len - 1] == 0)
		q.len--;
	// Zap any/all leading zeros in remainder
	zapLeadingZeros();
	// Deallocate temporary array.
	// (Thanks to Brad Spencer for noticing my accidental omission of this!)
	delete [] work2;
	
}
/*
* The out-of-bounds accesses story:
* 
* On 2005.01.06 or 2005.01.07 (depending on your time zone),
* Milan Tomic reported out-of-bounds memory accesses in
* the Big Integer Library.  To investigate the problem, I
* added code to bounds-check every access to the `blk' array
* of a `NumberlikeArray'.
*
* This gave me warnings that fell into two categories of false
* positives.  The bounds checker was based on length, not
* capacity, and in two places I had accessed memory that I knew
* was inside the capacity but that wasn't inside the length:
* 
* (1) The extra zero block at the left of `*this'.  Earlier
* versions said `allocateAndCopy(len + 1); blk[len] = 0;'
* but did not increment `len'.
*
* (2) The entire digit array in the conversion constructor
* ``BigUnsignedInABase(BigUnsigned)''.  It was allocated with
* a conservatively high capacity, but the length wasn't set
* until the end of the constructor.
*
* To simplify matters, I changed both sections of code so that
* all accesses occurred within the length.  The messages went
* away, and I told Milan that I couldn't reproduce the problem,
* sending a development snapshot of the bounds-checked code.
*
* Then, on 2005.01.09-10, he told me his debugger still found
* problems, specifically at the line `delete [] work2'.
* It was `work2', not `blk', that was causing the problems;
* this possibility had not occurred to me at all.  In fact,
* the problem was that `work2' needed an extra block just
* like `*this'.  Go ahead and laugh at me for finding (1)
* without seeing what was actually causing the trouble.  :-)
*
* The 2005.01.11 version fixes this problem.  I hope this is
* the last of my memory-related bloopers.  So this is what
* starts happening to your C++ code if you use Java too much!
*/

// Bitwise and
void BigUnsigned::bitAnd(const BigUnsigned &a, const BigUnsigned &b) {
	DTRT_ALIASED(this == &a || this == &b, bitAnd(a, b));
	len = (a.len >= b.len) ? b.len : a.len;
	allocate(len);
	Index i;
	for (i = 0; i < len; i++)
		blk[i] = a.blk[i] & b.blk[i];
	zapLeadingZeros();
}

// Bitwise or
void BigUnsigned::bitOr(const BigUnsigned &a, const BigUnsigned &b) {
	DTRT_ALIASED(this == &a || this == &b, bitOr(a, b));
	Index i;
	const BigUnsigned *a2, *b2;
	if (a.len >= b.len) {
		a2 = &a;
		b2 = &b;
	} else {
		a2 = &b;
		b2 = &a;
	}
	allocate(a2->len);
	for (i = 0; i < b2->len; i++)
		blk[i] = a2->blk[i] | b2->blk[i];
	for (; i < a2->len; i++)
		blk[i] = a2->blk[i];
	len = a2->len;
}

// Bitwise xor
void BigUnsigned::bitXor(const BigUnsigned &a, const BigUnsigned &b) {
	DTRT_ALIASED(this == &a || this == &b, bitXor(a, b));
	Index i;
	const BigUnsigned *a2, *b2;
	if (a.len >= b.len) {
		a2 = &a;
		b2 = &b;
	} else {
		a2 = &b;
		b2 = &a;
	}
	allocate(a2->len);
	for (i = 0; i < b2->len; i++)
		blk[i] = a2->blk[i] ^ b2->blk[i];
	for (; i < a2->len; i++)
		blk[i] = a2->blk[i];
	len = a2->len;
	zapLeadingZeros();
}

// Bitwise shift left
void BigUnsigned::bitShiftLeft(const BigUnsigned &a, unsigned int b) {
	DTRT_ALIASED(this == &a, bitShiftLeft(a, b));
	Index shiftBlocks = b / N;
	unsigned int shiftBits = b % N;
	// + 1: room for high bits nudged left into another block
	len = a.len + shiftBlocks + 1;
	allocate(len);
	Index i, j;
	for (i = 0; i < shiftBlocks; i++)
		blk[i] = 0;
	for (j = 0, i = shiftBlocks; j <= a.len; j++, i++)
		blk[i] = getShiftedBlock(a, j, shiftBits);
	// Zap possible leading zero
	if (blk[len - 1] == 0)
		len--;
}

// Bitwise shift right
void BigUnsigned::bitShiftRight(const BigUnsigned &a, unsigned int b) {
	DTRT_ALIASED(this == &a, bitShiftRight(a, b));
	// This calculation is wacky, but expressing the shift as a left bit shift
	// within each block lets us use getShiftedBlock.
	Index rightShiftBlocks = (b + N - 1) / N;
	unsigned int leftShiftBits = N * rightShiftBlocks - b;
	// Now (N * rightShiftBlocks - leftShiftBits) == b
	// and 0 <= leftShiftBits < N.
	if (rightShiftBlocks >= a.len + 1) {
		// All of a is guaranteed to be shifted off, even considering the left
		// bit shift.
		len = 0;
		return;
	}
	// Now we're allocating a positive amount.
	// + 1: room for high bits nudged left into another block
	len = a.len + 1 - rightShiftBlocks;
	allocate(len);
	Index i, j;
	for (j = rightShiftBlocks, i = 0; j <= a.len; j++, i++)
		blk[i] = getShiftedBlock(a, j, leftShiftBits);
	// Zap possible leading zero
	if (blk[len - 1] == 0)
		len--;
}

// INCREMENT/DECREMENT OPERATORS

// Prefix increment
void BigUnsigned::operator ++() {
	Index i;
	bool carry = true;
	for (i = 0; i < len && carry; i++) {
		blk[i]++;
		carry = (blk[i] == 0);
	}
	if (carry) {
		// Matt fixed a bug 2004.12.24: next 2 lines used to say allocateAndCopy(len + 1)
		// Matt fixed another bug 2006.04.24:
		// old number only has len blocks, so copy before increasing length
		allocateAndCopy(len + 1);
		len++;
		blk[i] = 1;
	}
}

// Postfix increment: same as prefix
void BigUnsigned::operator ++(int) {
	operator ++();
}

// Prefix decrement
void BigUnsigned::operator --() {
	if (len == 0)
		throw "BigUnsigned::operator --(): Cannot decrement an unsigned zero";
	Index i;
	bool borrow = true;
	for (i = 0; borrow; i++) {
		borrow = (blk[i] == 0);
		blk[i]--;
	}
	// Zap possible leading zero (there can only be one)
	if (blk[len - 1] == 0)
		len--;
}

// Postfix decrement: same as prefix
void BigUnsigned::operator --(int) {
	operator --();
}