#include "BigUnsigned.hh" // The "management" routines that used to be here are now in NumberlikeArray.hh. /* * The steps for construction of a BigUnsigned * from an integral value x are as follows: * 1. If x is zero, create an empty BigUnsigned and stop. * 2. If x is negative, throw an exception. * 3. Allocate a one-block number array. * 4. If x is of a signed type, convert x to the unsigned * type of the same length. * 5. Expand x to a Blk, and store it in the number array. * * Since 2005.01.06, NumberlikeArray uses `NULL' rather * than a real array if one of zero length is needed. * These constructors implicitly call NumberlikeArray's * default constructor, which sets `blk = NULL, cap = len = 0'. * So if the input number is zero, they can just return. * See remarks in `NumberlikeArray.hh'. */ BigUnsigned::BigUnsigned(unsigned long x) { if (x == 0) ; // NumberlikeArray already did all the work else { cap = 1; blk = new Blk[1]; len = 1; blk[0] = Blk(x); } } BigUnsigned::BigUnsigned(long x) { if (x == 0) ; else if (x > 0) { cap = 1; blk = new Blk[1]; len = 1; blk[0] = Blk(x); } else throw "BigUnsigned::BigUnsigned(long): Cannot construct a BigUnsigned from a negative number"; } BigUnsigned::BigUnsigned(unsigned int x) { if (x == 0) ; else { cap = 1; blk = new Blk[1]; len = 1; blk[0] = Blk(x); } } BigUnsigned::BigUnsigned(int x) { if (x == 0) ; else if (x > 0) { cap = 1; blk = new Blk[1]; len = 1; blk[0] = Blk(x); } else throw "BigUnsigned::BigUnsigned(int): Cannot construct a BigUnsigned from a negative number"; } BigUnsigned::BigUnsigned(unsigned short x) { if (x == 0) ; else { cap = 1; blk = new Blk[1]; len = 1; blk[0] = Blk(x); } } BigUnsigned::BigUnsigned(short x) { if (x == 0) ; else if (x > 0) { cap = 1; blk = new Blk[1]; len = 1; blk[0] = Blk(x); } else throw "BigUnsigned::BigUnsigned(short): Cannot construct a BigUnsigned from a negative number"; } // CONVERTERS /* * The steps for conversion of a BigUnsigned to an * integral type are as follows: * 1. If the BigUnsigned is zero, return zero. * 2. If it is more than one block long or its lowest * block has bits set out of the range of the target * type, throw an exception. * 3. Otherwise, convert the lowest block to the * target type and return it. */ namespace { // These masks are used to test whether a Blk has bits // set out of the range of a smaller integral type. Note // that this range is not considered to include the sign bit. const BigUnsigned::Blk lMask = ~0 >> 1; const BigUnsigned::Blk uiMask = (unsigned int)(~0); const BigUnsigned::Blk iMask = uiMask >> 1; const BigUnsigned::Blk usMask = (unsigned short)(~0); const BigUnsigned::Blk sMask = usMask >> 1; } BigUnsigned::operator unsigned long() const { if (len == 0) return 0; else if (len == 1) return (unsigned long) blk[0]; else throw "BigUnsigned::operator unsigned long: Value is too big for an unsigned long"; } BigUnsigned::operator long() const { if (len == 0) return 0; else if (len == 1 && (blk[0] & lMask) == blk[0]) return (long) blk[0]; else throw "BigUnsigned::operator long: Value is too big for a long"; } BigUnsigned::operator unsigned int() const { if (len == 0) return 0; else if (len == 1 && (blk[0] & uiMask) == blk[0]) return (unsigned int) blk[0]; else throw "BigUnsigned::operator unsigned int: Value is too big for an unsigned int"; } BigUnsigned::operator int() const { if (len == 0) return 0; else if (len == 1 && (blk[0] & iMask) == blk[0]) return (int) blk[0]; else throw "BigUnsigned::operator int: Value is too big for an int"; } BigUnsigned::operator unsigned short() const { if (len == 0) return 0; else if (len == 1 && (blk[0] & usMask) == blk[0]) return (unsigned short) blk[0]; else throw "BigUnsigned::operator unsigned short: Value is too big for an unsigned short"; } BigUnsigned::operator short() const { if (len == 0) return 0; else if (len == 1 && (blk[0] & sMask) == blk[0]) return (short) blk[0]; else throw "BigUnsigned::operator short: Value is too big for a short"; } // COMPARISON BigUnsigned::CmpRes BigUnsigned::compareTo(const BigUnsigned &x) const { // A bigger length implies a bigger number. if (len < x.len) return less; else if (len > x.len) return greater; else { // Compare blocks one by one from left to right. Index i = len; while (i > 0) { i--; if (blk[i] == x.blk[i]) continue; else if (blk[i] > x.blk[i]) return greater; else return less; } // If no blocks differed, the numbers are equal. return equal; } } // PUT-HERE OPERATIONS /* * Below are implementations of the four basic arithmetic operations * for `BigUnsigned's. Their purpose is to use a mechanism that can * calculate the sum, difference, product, and quotient/remainder of * two individual blocks in order to calculate the sum, difference, * product, and quotient/remainder of two multi-block BigUnsigned * numbers. * * As alluded to in the comment before class `BigUnsigned', * these algorithms bear a remarkable similarity (in purpose, if * not in implementation) to the way humans operate on big numbers. * The built-in `+', `-', `*', `/' and `%' operators are analogous * to elementary-school ``math facts'' and ``times tables''; the * four routines below are analogous to ``long division'' and its * relatives. (Only a computer can ``memorize'' a times table with * 18446744073709551616 entries! (For 32-bit blocks.)) * * The discovery of these four algorithms, called the ``classical * algorithms'', marked the beginning of the study of computer science. * See Section 4.3.1 of Knuth's ``The Art of Computer Programming''. */ /* * On most calls to put-here operations, it's safe to read the inputs little by * little and write the outputs little by little. However, if one of the * inputs is coming from the same variable into which the output is to be * stored (an "aliased" call), we risk overwriting the input before we read it. * In this case, we first compute the result into a temporary BigUnsigned * variable and then copy it into the requested output variable *this. * Each put-here operation uses the DTRT_ALIASED macro (Do The Right Thing on * aliased calls) to generate code for this check. * * I adopted this approach on 2007.02.13 (see Assignment Operators in * BigUnsigned.hh). Before then, put-here operations rejected aliased calls * with an exception. I think doing the right thing is better. * * Some of the put-here operations can probably handle aliased calls safely * without the extra copy because (for example) they process blocks strictly * right-to-left. At some point I might determine which ones don't need the * copy, but my reasoning would need to be verified very carefully. For now * I'll leave in the copy. */ #define DTRT_ALIASED(cond, op) \ if (cond) { \ BigUnsigned tmpThis; \ tmpThis.op; \ *this = tmpThis; \ return; \ } // Addition void BigUnsigned::add(const BigUnsigned &a, const BigUnsigned &b) { DTRT_ALIASED(this == &a || this == &b, add(a, b)); // If one argument is zero, copy the other. if (a.len == 0) { operator =(b); return; } else if (b.len == 0) { operator =(a); return; } // Some variables... // Carries in and out of an addition stage bool carryIn, carryOut; Blk temp; Index i; // a2 points to the longer input, b2 points to the shorter const BigUnsigned *a2, *b2; if (a.len >= b.len) { a2 = &a; b2 = &b; } else { a2 = &b; b2 = &a; } // Set prelimiary length and make room in this BigUnsigned len = a2->len + 1; allocate(len); // For each block index that is present in both inputs... for (i = 0, carryIn = false; i < b2->len; i++) { // Add input blocks temp = a2->blk[i] + b2->blk[i]; // If a rollover occurred, the result is less than either input. // This test is used many times in the BigUnsigned code. carryOut = (temp < a2->blk[i]); // If a carry was input, handle it if (carryIn) { temp++; carryOut |= (temp == 0); } blk[i] = temp; // Save the addition result carryIn = carryOut; // Pass the carry along } // If there is a carry left over, increase blocks until // one does not roll over. for (; i < a2->len && carryIn; i++) { temp = a2->blk[i] + 1; carryIn = (temp == 0); blk[i] = temp; } // If the carry was resolved but the larger number // still has blocks, copy them over. for (; i < a2->len; i++) blk[i] = a2->blk[i]; // Set the extra block if there's still a carry, decrease length otherwise if (carryIn) blk[i] = 1; else len--; } // Subtraction void BigUnsigned::subtract(const BigUnsigned &a, const BigUnsigned &b) { DTRT_ALIASED(this == &a || this == &b, subtract(a, b)); // If b is zero, copy a. If a is shorter than b, the result is negative. if (b.len == 0) { operator =(a); return; } else if (a.len < b.len) throw "BigUnsigned::subtract: Negative result in unsigned calculation"; // Some variables... bool borrowIn, borrowOut; Blk temp; Index i; // Set preliminary length and make room len = a.len; allocate(len); // For each block index that is present in both inputs... for (i = 0, borrowIn = false; i < b.len; i++) { temp = a.blk[i] - b.blk[i]; // If a reverse rollover occurred, the result is greater than the block from a. borrowOut = (temp > a.blk[i]); // Handle an incoming borrow if (borrowIn) { borrowOut |= (temp == 0); temp--; } blk[i] = temp; // Save the subtraction result borrowIn = borrowOut; // Pass the borrow along } // If there is a borrow left over, decrease blocks until // one does not reverse rollover. for (; i < a.len && borrowIn; i++) { borrowIn = (a.blk[i] == 0); blk[i] = a.blk[i] - 1; } // If there's still a borrow, the result is negative. // Throw an exception, but zero out this object first just in case. if (borrowIn) { len = 0; throw "BigUnsigned::subtract: Negative result in unsigned calculation"; } else // Copy over the rest of the blocks for (; i < a.len; i++) blk[i] = a.blk[i]; // Zap leading zeros zapLeadingZeros(); } /* * About the multiplication and division algorithms: * * I searched unsucessfully for fast built-in operations like the `b_0' * and `c_0' Knuth describes in Section 4.3.1 of ``The Art of Computer * Programming'' (replace `place' by `Blk'): * * ``b_0[:] multiplication of a one-place integer by another one-place * integer, giving a two-place answer; * * ``c_0[:] division of a two-place integer by a one-place integer, * provided that the quotient is a one-place integer, and yielding * also a one-place remainder.'' * * I also missed his note that ``[b]y adjusting the word size, if * necessary, nearly all computers will have these three operations * available'', so I gave up on trying to use algorithms similar to his. * A future version of the library might include such algorithms; I * would welcome contributions from others for this. * * I eventually decided to use bit-shifting algorithms. To multiply `a' * and `b', we zero out the result. Then, for each `1' bit in `a', we * shift `b' left the appropriate amount and add it to the result. * Similarly, to divide `a' by `b', we shift `b' left varying amounts, * repeatedly trying to subtract it from `a'. When we succeed, we note * the fact by setting a bit in the quotient. While these algorithms * have the same O(n^2) time complexity as Knuth's, the ``constant factor'' * is likely to be larger. * * Because I used these algorithms, which require single-block addition * and subtraction rather than single-block multiplication and division, * the innermost loops of all four routines are very similar. Study one * of them and all will become clear. */ /* * This is a little inline function used by both the multiplication * routine and the division routine. * * `getShiftedBlock' returns the `x'th block of `num << y'. * `y' may be anything from 0 to N - 1, and `x' may be anything from * 0 to `num.len'. * * Two things contribute to this block: * * (1) The `N - y' low bits of `num.blk[x]', shifted `y' bits left. * * (2) The `y' high bits of `num.blk[x-1]', shifted `N - y' bits right. * * But we must be careful if `x == 0' or `x == num.len', in * which case we should use 0 instead of (2) or (1), respectively. * * If `y == 0', then (2) contributes 0, as it should. However, * in some computer environments, for a reason I cannot understand, * `a >> b' means `a >> (b % N)'. This means `num.blk[x-1] >> (N - y)' * will return `num.blk[x-1]' instead of the desired 0 when `y == 0'; * the test `y == 0' handles this case specially. */ inline BigUnsigned::Blk getShiftedBlock(const BigUnsigned &num, BigUnsigned::Index x, unsigned int y) { BigUnsigned::Blk part1 = (x == 0 || y == 0) ? 0 : (num.blk[x - 1] >> (BigUnsigned::N - y)); BigUnsigned::Blk part2 = (x == num.len) ? 0 : (num.blk[x] << y); return part1 | part2; } // Multiplication void BigUnsigned::multiply(const BigUnsigned &a, const BigUnsigned &b) { DTRT_ALIASED(this == &a || this == &b, multiply(a, b)); // If either a or b is zero, set to zero. if (a.len == 0 || b.len == 0) { len = 0; return; } /* * Overall method: * * Set this = 0. * 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 --(); }