[libc] Optimize BigInt→decimal in IntegerToString

When IntegerToString converts a BigInt into decimal, it determines
each digit by computing `n % 10` and then resets n to `n / 10`, until
the number becomes zero. The div and mod operations are done using
`BigInt::divide_unsigned`, which uses the simplest possible bit-by-bit
iteration, which is a slow algorithm in general, but especially so if
the divisor 10 must first be promoted to a BigInt the same size as the
dividend. The effect is to make each division take quadratic time, so
that the overall decimal conversion is cubic – and the division is
quadratic in the number of _bits_, so the constant of proportionality
is also large.

In this patch I've provided custom code to extract decimal digits much
faster, based on knowing that the divisor is always 10, and processing
a word at a time. So each digit extraction is linear-time with a much
smaller constant of proportionality.

Full comments are in the code. The general strategy is to do the
reduction mod 10 first to determine the output digit; then subtract it
off, so that what's left is guaranteed to be an exact multiple of 10;
and finally divide by 10 using modular-arithmetic techniques rather
than reciprocal-approximation-based ordinary integer division.

I didn't find any existing tests of IntegerToString on a BigInt, so
I've added one.

Author: statham-arm

libc/src/__support/integer_to_string.h

@@ -164,6 +164,168 @@ template <size_t radix> using Custom = details::Fmt<radix>;
 
 } // namespace radix
 
+// Extract the low-order decimal digit from a value of integer type T. The
+// returned value is the digit itself, from 0 to 9. The input value is passed
+// by reference, and modified by dividing by 10, so that iterating this
+// function extracts all the digits of the original number one at a time from
+// low to high.
+template <typename T, cpp::enable_if_t<cpp::is_integral_v<T>, int> = 0>
+LIBC_INLINE uint8_t extract_decimal_digit(T &value) {
+  const uint8_t digit(static_cast<uint8_t>(value % 10));
+  // For built-in integer types, we assume that an adequately fast division is
+  // available. If hardware division isn't implemented, then with a divisor
+  // known at compile time the compiler might be able to generate an optimized
+  // sequence instead.
+  value /= 10;
+  return digit;
+}
+
+// A specialization of extract_decimal_digit for the BigInt type in big_int.h,
+// avoiding the use of general-purpose BigInt division which is very slow.
+template <typename T, cpp::enable_if_t<is_big_int_v<T>, int> = 0>
+LIBC_INLINE uint8_t extract_decimal_digit(T &value) {
+  // There are two essential ways you can turn n into (n/10,n%10). One is
+  // ordinary integer division. The other is a modular-arithmetic approach in
+  // which you first compute n%10 by bit twiddling, then subtract it off to get
+  // a value that is definitely a multiple of 10. Then you divide that by 10 in
+  // two steps: shift right to divide off a factor of 2, and then divide off a
+  // factor of 5 by multiplying by the modular inverse of 5 mod 2^BITS. (That
+  // last step only works if you know there's no remainder, which is why you
+  // had to subtract off the output digit first.)
+  //
+  // Either approach can be made to work in linear time. This code uses the
+  // modular-arithmetic technique, because the other approach either does a lot
+  // of integer divisions (requiring a fast hardware divider), or else uses a
+  // "multiply by an approximation to the reciprocal" technique which depends
+  // on careful error analysis which might go wrong in an untested edge case.
+
+  using Word = typename T::word_type;
+
+  // Find the remainder (value % 10). We do this by breaking up the input
+  // integer into chunks of size WORD_SIZE/2, so that the sum of them doesn't
+  // overflow a Word. Then we sum all the half-words times 6, except the bottom
+  // one, which is added to that sum without scaling.
+  //
+  // Why 6? Because you can imagine that the original number had the form
+  //
+  //   halfwords[0] + K*halfwords[1] + K^2*halfwords[2] + ...
+  //
+  // where K = 2^(WORD_SIZE/2). Since WORD_SIZE is expected to be a multiple of
+  // 8, that makes WORD_SIZE/2 a multiple of 4, so that K is a power of 16. And
+  // all powers of 16 (larger than 1) are congruent to 6 mod 10, by induction:
+  // 16 itself is, and 6^2=36 is also congruent to 6.
+  Word acc_remainder = 0;
+  const Word HALFWORD_BITS = T::WORD_SIZE / 2;
+  const Word HALFWORD_MASK = ((Word(1) << HALFWORD_BITS) - 1);
+  // Sum both halves of all words except the low one.
+  for (size_t i = 1; i < T::WORD_COUNT; i++) {
+    acc_remainder += value.val[i] >> HALFWORD_BITS;
+    acc_remainder += value.val[i] & HALFWORD_MASK;
+  }
+  // Add the high half of the low word. Then we have everything that needs to
+  // be multiplied by 6, so do that.
+  acc_remainder += value.val[0] >> HALFWORD_BITS;
+  acc_remainder *= 6;
+  // Having multiplied it by 6, add the lowest half-word, and then reduce mod
+  // 10 by normal integer division to finish.
+  acc_remainder += value.val[0] & HALFWORD_MASK;
+  uint8_t digit = acc_remainder % 10;
+
+  // Now we have the output digit. Subtract it from the input value, and shift
+  // right to divide by 2.
+  value -= digit;
+  value >>= 1;
+
+  // Now all that's left is to multiply by the inverse of 5 mod 2^BITS. No
+  // matter what the value of BITS, the inverse of 5 has the very convenient
+  // form 0xCCCC...CCCD, with as many C hex digits in the middle as necessary.
+  //
+  // We could construct a second BigInt with all words 0xCCCCCCCCCCCCCCCC,
+  // increment the bottom word, and call a general-purpose multiply function.
+  // But we can do better, by taking advantage of the regularity: we can do
+  // this particular operation in linear time, whereas a general multiplier
+  // would take superlinear time (quadratic in small cases).
+  //
+  // To begin with, instead of computing n*0xCCCC...CCCD, we'll compute
+  // n*0xCCCC...CCCC and then add it to the original n. Then all the words of
+  // the multiplier have the same value 0xCCCCCCCCCCCCCCCC, which I'll just
+  // denote as C. If we also write t = 2^WORD_SIZE, and imagine (as an example)
+  // that the input number has three words x,y,z with x being the low word,
+  // then we're computing
+  //
+  //   (x + y t + z t^2) * (C + C t + C t^2)
+  //
+  // = x C + y C t + z C t^2
+  //       + x C t + y C t^2 + z C t^3
+  //               + x C t^2 + y C t^3 + z C t^4
+  //
+  // but we're working mod t^3, so the high-order terms vanish and this becomes
+  //
+  //   x C + y C t + z C t^2
+  //       + x C t + y C t^2
+  //               + x C t^2
+  //
+  // = x C + (x+y) C t + (x+y+z) C t^2
+  //
+  // So all you have to do is to work from the low word of the integer upwards,
+  // accumulating C times the sum of all the words you've seen so far to get
+  // x*C, (x+y)*C, (x+y+z)*C and so on. In each step you add another product to
+  // the accumulator, and add the accumulator to the corresponding word of the
+  // original number (so that we end up with value*CCCD, not just value*CCCC).
+  //
+  // If you do that literally, then your accumulator has to be three words
+  // wide, because the sum of words can overflow into a second word, and
+  // multiplying by C adds another word. But we can do slightly better by
+  // breaking each product word*C up into a bottom half and a top half. If we
+  // write x*C = xl + xh*t, and similarly for y and z, then our sum becomes
+  //
+  //   (xl + xh t) + (yl + yh t) t + (zl + zh t) t^2
+  //               + (xl + xh t) t + (yl + yh t) t^2
+  //                               + (xl + xh t) t^2
+  //
+  // and if you expand out again, collect terms, and discard t^3 terms, you get
+  //
+  //   (xl)
+  // + (xl + xh + yl) t
+  // + (xl + xh + yl + yh + zl) t^2
+  //
+  // in which each coefficient is the sum of all the low words of the products
+  // up to _and including_ the current word, plus all the high words up to but
+  // _not_ including the current word. So now you only have to retain two words
+  // of sum instead of three.
+  //
+  // We do this entire procedure in a single in-place pass over the input
+  // number, reading each word to make its product with C and then adding the
+  // low word of the accumulator to it.
+  const Word C = (Word(0) - 1) / 5 * 4; // calculate 0xCCCC as 4/5 of 0xFFFF
+  Word acc_lo = 0, acc_hi = 0; // accumulator of all the half-products so far
+  Word carry_bit, carry_word = 0;
+
+  for (size_t i = 0; i < T::WORD_COUNT; i++) {
+    // Make the two-word product of C with the current input word.
+    multiword::DoubleWide<Word> product = multiword::mul2(C, value.val[i]);
+
+    // Add the low half of the product to our accumulator, but not yet the high
+    // half.
+    acc_lo = add_with_carry<Word>(acc_lo, product[0], 0, carry_bit);
+    acc_hi += carry_bit;
+
+    // Now the accumulator contains exactly the value we need to add to the
+    // current input word. Add it, plus any carries from lower words, and make
+    // a new word of carry data to propagate into the next iteration.
+    value.val[i] = add_with_carry<Word>(value.val[i], carry_word, 0, carry_bit);
+    carry_word = acc_hi + carry_bit;
+    value.val[i] = add_with_carry<Word>(value.val[i], acc_lo, 0, carry_bit);
+    carry_word += carry_bit;
+
+    // Now add the high half of the current product to our accumulator.
+    acc_lo = add_with_carry<Word>(acc_lo, product[1], 0, carry_bit);
+    acc_hi += carry_bit;
+  }
+
+  return digit;
+}
+
 // See file header for documentation.
 template <typename T, typename Fmt = radix::Dec> class IntegerToString {
   static_assert(cpp::is_integral_v<T> || is_big_int_v<T>);
@@ -229,6 +391,15 @@ template <typename T, typename Fmt = radix::Dec> class IntegerToString {
       }
     }
 
+    LIBC_INLINE static void
+    write_unsigned_number_dec(UNSIGNED_T value,
+                              details::BackwardStringBufferWriter &sink) {
+      while (sink.ok() && value != 0) {
+        const uint8_t digit = extract_decimal_digit(value);
+        sink.push(digit_char(digit));
+      }
+    }
+
     // Returns the absolute value of 'value' as 'UNSIGNED_T'.
     LIBC_INLINE static UNSIGNED_T abs(T value) {
       if (cpp::is_unsigned_v<T> || value >= 0)
@@ -256,7 +427,7 @@ template <typename T, typename Fmt = radix::Dec> class IntegerToString {
     LIBC_INLINE static void write(T value,
                                   details::BackwardStringBufferWriter &sink) {
       if constexpr (Fmt::BASE == 10) {
-        write_unsigned_number(abs(value), sink);
+        write_unsigned_number_dec(abs(value), sink);
       } else {
         write_unsigned_number(static_cast<UNSIGNED_T>(value), sink);
       }

libc/test/src/__support/integer_to_string_test.cpp

@@ -16,6 +16,7 @@
 
 #include "test/UnitTest/Test.h"
 
+using LIBC_NAMESPACE::BigInt;
 using LIBC_NAMESPACE::IntegerToString;
 using LIBC_NAMESPACE::cpp::span;
 using LIBC_NAMESPACE::cpp::string_view;
@@ -297,6 +298,51 @@ TEST(LlvmLibcIntegerToStringTest, Sign) {
   EXPECT(DEC, 1, "+1");
 }
 
+TEST(LlvmLibcIntegerToStringTest, BigInt_Base_10) {
+  uint64_t uint256_max[4] = {
+      0xFFFFFFFFFFFFFFFF,
+      0xFFFFFFFFFFFFFFFF,
+      0xFFFFFFFFFFFFFFFF,
+      0xFFFFFFFFFFFFFFFF,
+  };
+  uint64_t int256_max[4] = {
+      0xFFFFFFFFFFFFFFFF,
+      0xFFFFFFFFFFFFFFFF,
+      0xFFFFFFFFFFFFFFFF,
+      0x7FFFFFFFFFFFFFFF,
+  };
+  uint64_t int256_min[4] = {
+      0,
+      0,
+      0,
+      0x8000000000000000,
+  };
+
+  using unsigned_type = IntegerToString<BigInt<256, false>, Dec>;
+  EXPECT(unsigned_type, 0, "0");
+  EXPECT(unsigned_type, 1, "1");
+  EXPECT(unsigned_type, uint256_max,
+         "115792089237316195423570985008687907853269984665640564039457584007913"
+         "129639935");
+  EXPECT(unsigned_type, int256_max,
+         "578960446186580977117854925043439539266349923328202820197287920039565"
+         "64819967");
+  EXPECT(unsigned_type, int256_min,
+         "578960446186580977117854925043439539266349923328202820197287920039565"
+         "64819968");
+
+  using signed_type = IntegerToString<BigInt<256, true>, Dec>;
+  EXPECT(signed_type, 0, "0");
+  EXPECT(signed_type, 1, "1");
+  EXPECT(signed_type, uint256_max, "-1");
+  EXPECT(signed_type, int256_max,
+         "578960446186580977117854925043439539266349923328202820197287920039565"
+         "64819967");
+  EXPECT(signed_type, int256_min,
+         "-57896044618658097711785492504343953926634992332820282019728792003956"
+         "564819968");
+}
+
 TEST(LlvmLibcIntegerToStringTest, BufferOverrun) {
   { // Writing '0' in an empty buffer requiring zero digits : works
     const auto view =