Donald Knuth’s Algorithm D
, its implementation in Hacker’s Delight
, and elsewhere
- Purpose
- Introduction
- Generic implementation in Hacker’s Delight
- Specialised implementation in Hacker’s Delight
- Flawed and poor implementation in libdivide
- Poor implementation in Free Pascal’s
Run-Time Library
- Poor implementation in Google’s Go Programming Language
- Implementation in Google’s V8 JavaScript engine
- Proper (and optimised) implementation …
- … in ANSI C
- … in i386 Assembler
Purpose
Show some shortcomings of the multiple precision division implementations presented in Hacker’s Delight, and their effect elsewhere.Introduction
In Volume 2: Seminumerical Algorithms, Chapter 4.3: Multiple-Precision Arithmetic of The Art of Computer Programming, Donald Ervin Knuth presentsAlgorithm D (Division of nonnegative integers):
Note: I added an introductory step D0 providing definitions and descriptions.
Algorithm D (Division of nonnegative integers).
- D0. [Define]
- Let U be the dividend (or numerator) of m+n
digits, stored in an array of m+n+1 elements, onedigitper element, with the most significantdigitin element U[m+n−1] and the least significantdigitin element U[0];- Let V be the divisor (or denominator) of n
digits, stored in a second array of n elements, with n greater than 1, a non-zero most significantdigitin element V[n−1] and the least significantdigitin element V[0];- Let B be the base (or radix) of the
digits(alsolimbs,placesorwords), typically (a power of) a power of 2, with B greater than 1.- (The algorithm computes the quotient Q of m+1
digitsas ⌊U ⁄ V⌋ (also U div V or U ÷ V), and the remainder R of ndigitsas U modulo V (also U mod V or U % V), using the followingprimitiveoperations:
¹ addition or subtraction of two single-digit integers, giving a single-digit sum and a carry, or a single-digit difference and a borrow;
²wideningmultiplication of two single-digit integers, giving a double-digit product;
³narrowingdivision of a double-digit integer by a single-digit integer, giving a single-digit quotient and a single-digit remainder.)- D1. [Normalize]
- Set D to (B − 1) ÷ V[n−1];
- Multiply all
digitsof U and V by D.- (On a binary computer, choose D to be a power of 2 instead of the value provided above; any value of D that results in V[n−1] not less than B ÷ 2 will suffice. If D is greater than 1, the eventual overflow
digitof the dividend U goes into element U[m+n].)- D2. [Initialize j]
- Set the loop counter j to m.
- D3. [Calculate Q′]
- Set Q′ to (U[n+j] * B + U[n−1+j]) ÷ V[n−1];
- Set R′ to (U[n+j] * B + U[n−1+j]) % V[n−1];
- Test if Q′ equals B or Q′ * V[n−2] is greater than R′ * B + U[n−2+j];
- If yes, then decrease Q′ by 1, increase R′ by V[n−1], and repeat this test while R is less than B.
- D4. [Multiply and subtract]
- Replace (U[n+j]U[n−1+j]…U[j]) by (U[n+j]U[n−1+j]…U[j]) − Q′ * (V[n−1]…V[1]V[0]).
- (The
digits(U[n+j]…U[j]) should be kept positive; if the result of this step is actually negative, (U[n+j]…U[j]) should be left as the true value plus Bn+1, namely as the B’s complement of the true value, and a borrow to the left should be remembered.)- D5. [Test remainder]
- Set Q[j] to Q′;
- If the result of step D4 was negative, i.e. the subtraction needed a borrow, then proceed with step D6; otherwise proceed with step D7.
- D6. [Add back]
- Decrease Q[j] by 1 and add (0V[n−1]…V[1]V[0]) to (U[n+j]U[n−1+j]…U[1+j]U[j]).
- (A carry will occur to the left of U[n+j], and it should be ignored since it cancels with the borrow that occurred in step D4.)
- D7. [Loop on j]
- Decrease j by 1;
- Test if j is not less than 0;
- If yes, go back to step D3.
- D8. [Unnormalize]
- Now (Q[m]…Q[1]Q[0]) is the desired quotient Q, and the desired remainder R may be obtained by dividing (U[n−1]…U[1]U[0]) by D.
Generic implementation in Hacker’s Delight
In his book Hacker’s Delight, Henry S. ("Hank") Warren presents the following generic implementation divmnu64.c of Donald Knuth’sAlgorithm Din ANSI C.
Note: I removed some parts of the code which are not relevant here.
/* This divides an n-word dividend by an m-word divisor, giving an
n-m+1-word quotient and m-word remainder. The bignums are in arrays of
words. Here a "word" is 32 bits. This routine is designed for a 64-bit
machine which has a 64/64 division instruction. */ 1
…
/* q[0], r[0], u[0], and v[0] contain the LEAST significant words.
(The sequence is in little-endian order).
This is a fairly precise implementation of Knuth's Algorithm D, for a
binary computer with base b = 2**32. The caller supplies:
1. Space q for the quotient, m - n + 1 words (at least one).
2. Space r for the remainder (optional), n words.
3. The dividend u, m words, m >= 1.
4. The divisor v, n words, n >= 2.
The most significant digit of the divisor, v[n-1], must be nonzero. The
dividend u may have leading zeros; this just makes the algorithm take
longer and makes the quotient contain more leading zeros. A value of
NULL may be given for the address of the remainder to signify that the
caller does not want the remainder.
The program does not alter the input parameters u and v.
The quotient and remainder returned may have leading zeros. The
function itself returns a value of 0 for success and 1 for invalid
parameters (e.g., division by 0).
For now, we must have m >= n. Knuth's Algorithm D also requires
that the dividend be at least as long as the divisor. (In his terms,
m >= 0 (unstated). Therefore m+n >= n.) */
int divmnu(unsigned q[], unsigned r[],
const unsigned u[], const unsigned v[],
int m, int n) {
const unsigned long long b = 4294967296LL; // Number base (2**32).
unsigned *un, *vn; // Normalized form of u, v.
unsigned long long qhat; // Estimated quotient digit.
unsigned long long rhat; // A remainder.
unsigned long long p; // Product of two digits.
long long t, k;
int s, i, j;
if (m < n || n <= 1 || v[n-1] == 0)
return 1; // Return if invalid param.
…
/* Normalize by shifting v left just enough so that its high-order
bit is on, and shift u left the same amount. We may have to append a
high-order digit on the dividend; we do that unconditionally. */
s = nlz(v[n-1]); // 0 <= s <= 31.
vn = (unsigned *)alloca(4*n);
for (i = n - 1; i > 0; i--)
vn[i] = (v[i] << s) | ((unsigned long long)v[i-1] >> (32-s));
vn[0] = v[0] << s;
un = (unsigned *)alloca(4*(m + 1));
un[m] = (unsigned long long)u[m-1] >> (32-s);
for (i = m - 1; i > 0; i--)
un[i] = (u[i] << s) | ((unsigned long long)u[i-1] >> (32-s));
un[0] = u[0] << s;
for (j = m - n; j >= 0; j--) { // Main loop.
// Compute estimate qhat of q[j].
qhat = (un[j+n]*b + un[j+n-1])/vn[n-1];
#ifdef OPTIMIZE
rhat = (un[j+n]*b + un[j+n-1])%vn[n-1];
#else // ORIGINAL
2 rhat = (un[j+n]*b + un[j+n-1]) - qhat*vn[n-1];
#endif
again:
if (qhat >= b ||
#ifdef OPTIMIZE
(unsigned)qhat*(unsigned long long)vn[n-2] > b*rhat + un[j+n-2]) {
#else // ORIGINAL
3 qhat*vn[n-2] > b*rhat + un[j+n-2]) {
#endif
qhat = qhat - 1;
rhat = rhat + vn[n-1];
if (rhat < b) goto again;
}
// Multiply and subtract.
k = 0;
for (i = 0; i < n; i++) {
#ifdef OPTIMIZE
p = (unsigned)qhat*(unsigned long long)vn[i];
#else // ORIGINAL
3 p = qhat*vn[i];
#endif
t = un[i+j] - k - (p & 0xFFFFFFFFLL);
un[i+j] = t;
k = (p >> 32) - (t >> 32);
}
t = un[j+n] - k;
un[j+n] = t;
q[j] = qhat; // Store quotient digit.
if (t < 0) { // If we subtracted too
q[j] = q[j] - 1; // much, add back.
k = 0;
for (i = 0; i < n; i++) {
t = (unsigned long long)un[i+j] + vn[i] + k;
un[i+j] = t;
k = t >> 32;
}
un[j+n] = un[j+n] + k;
}
} // End j.
// If the caller wants the remainder, unnormalize
// it and pass it back.
if (r != NULL) {
for (i = 0; i < n-1; i++)
r[i] = (un[i] >> s) | ((unsigned long long)un[i+1] << (32-s));
r[n-1] = un[n-1] >> s;
}
return 0;
}
…- Contrary to the highlighted part of its initial comment, this routine does not need a 64-bit machine; it but needs support for 64-bit integers and elementary 64-bit arithmetic operations, which ANSI C compilers provide since the last millennium on 32-bit machines too.
- Although a
64/64 division instruction
is explicitly stated, the code doesn’t take full advantage of it: instead to use the%operator tocomputefetch the remainder of the 64÷64-bit division (which comesfor free
on 64-bit machines, and almost for free with software implementations on 32-bit machines), it but performs an extraneous 64×64-bit multiplication — which is not free, especially on 32-bit machines! - The previous argument also holds for the multiplications
qhat*vn[n-2]andqhat*vn[i]: while an optimising compiler should detect that these are 32×32-bit multiplications returning a 64-bit product, I would not count on it — better give the compiler the proper hints!
Algorithm Dbuilds upon a 64÷32-bit division returning a 32-bit quotient and a 32-bit remainder, also known as
narrowingdivision, and a 32×32-bit multiplication returning a 64-bit product, also known as
wideningmultiplication, which both are but not supported in ANSI C!
Specialised implementation in Hacker’s Delight
Hank also presents a simplified and specialised implementation divlu.c for unsigned 64÷32-bit division, based on a 32÷16-bit division returning a 32-bit quotient and a 16-bit remainder, i.e. suited for 16-bit machines.Note: I removed some parts of the code which are not relevant here.
/* Long division, unsigned (64/32 ==> 32).
This procedure performs unsigned "long division" i.e., division of a
64-bit unsigned dividend by a 32-bit unsigned divisor, producing a
32-bit quotient. In the overflow cases (divide by 0, or quotient
exceeds 32 bits), it returns a remainder of 0xFFFFFFFF (an impossible
value).
The dividend is u1 and u0, with u1 being the most significant word.
The divisor is parameter v. The value returned is the quotient.
[…] Several of the variables below could be
"short," but having them fullwords gives better code on gcc/Intel.
[…]
This is the version that's in the hacker book. */
unsigned divlu2(unsigned u1, unsigned u0, unsigned v,
unsigned *r) {
const unsigned b = 65536; // Number base (16 bits).
unsigned un1, un0, // Norm. dividend LSD's.
vn1, vn0, // Norm. divisor digits.
q1, q0, // Quotient digits.
un32, un21, un10,// Dividend digit pairs.
rhat; // A remainder.
int s; // Shift amount for norm.
if (u1 >= v) { // If overflow, set rem.
if (r != NULL) // to an impossible value,
*r = 0xFFFFFFFF; // and return the largest
return 0xFFFFFFFF;} // possible quotient.
s = nlz(v); // 0 <= s <= 31.
v = v << s; // Normalize divisor.
vn1 = v >> 16; // Break divisor up into
vn0 = v & 0xFFFF; // two 16-bit digits.
un32 = (u1 << s) | (u0 >> 32 - s) & (-s >> 31);
un10 = u0 << s; // Shift dividend left.
un1 = un10 >> 16; // Break right half of
un0 = un10 & 0xFFFF; // dividend into two digits.
q1 = un32/vn1; // Compute the first
#ifdef OPTIMIZE
rhat = un32%vn1; // quotient digit, q1.
#else // ORIGINAL
rhat = un32 - q1*vn1; // quotient digit, q1.
#endif
again1:
if (q1 >= b || q1*vn0 > b*rhat + un1) {
q1 = q1 - 1;
rhat = rhat + vn1;
if (rhat < b) goto again1;}
un21 = un32*b + un1 - q1*v; // Multiply and subtract.
q0 = un21/vn1; // Compute the second
#ifdef OPTIMIZE
rhat = un21%vn1; // quotient digit, q0.
#else // ORIGINAL
rhat = un21 - q0*vn1; // quotient digit, q0.
#endif
again2:
if (q0 >= b || q0*vn0 > b*rhat + un0) {
q0 = q0 - 1;
rhat = rhat + vn1;
if (rhat < b) goto again2;}
if (r != NULL) // If remainder is wanted,
*r = (un21*b + un0 - q0*v) >> s; // return it.
return q1*b + q0;
}
…Flawed implementation in libdivide
Now take a look at the naïve and ratherWarning: don’t use this function; it demonstrates that its author copyist does not understand the original code and fails to read or ignores the comment, it exhibits poor performance, and it has a serious bug!
// libdivide.h
// Copyright 2010 - 2018 ridiculous_fish
…
// Code taken from Hacker's Delight:
// http://www.hackersdelight.org/HDcode/divlu.c.
// License permits inclusion here per:
// http://www.hackersdelight.org/permissions.htm
static uint64_t libdivide_128_div_64_to_64(uint64_t u1, uint64_t u0, uint64_t v, uint64_t *r) {
1 const uint64_t b = (1ULL << 32); // Number base (16 bits)
2 uint64_t un1, un0; // Norm. dividend LSD's
2 uint64_t vn1, vn0; // Norm. divisor digits
2 uint64_t q1, q0; // Quotient digits
3 uint64_t un64, un21, un10; // Dividend digit pairs
uint64_t rhat; // A remainder
#ifdef OPTIMIZE
uint64_t qhat; // A quotient
uint64_t p; // A product
#endif
int32_t s; // Shift amount for norm
// If overflow, set rem. to an impossible value,
// and return the largest possible quotient
if (u1 >= v) {
if (r != NULL)
*r = (uint64_t) -1;
return (uint64_t) -1;
}
// count leading zeros
s = libdivide__count_leading_zeros64(v);
if (s > 0) {
// Normalize divisor
v = v << s;
5 un64 = (u1 << s) | ((u0 >> (64 - s)) & (-s >> 31));
4 un10 = u0 << s; // Shift dividend left
} else {
// Avoid undefined behavior
6 un64 = u1 | u0;
4 un10 = u0;
}
// Break divisor up into two 32-bit digits
7 vn1 = v >> 32;
7 vn0 = v & 0xFFFFFFFF;
// Break right half of dividend into two digits
7 un1 = un10 >> 32;
7 un0 = un10 & 0xFFFFFFFF;
// Compute the first quotient digit, q1
#ifdef OPTIMIZE
qhat = un64 / vn1;
rhat = un64 % vn1;
p = qhat * vn0;
while (qhat >= b || p > b * rhat + un1) {
qhat = qhat - 1;
rhat = rhat + vn1;
if (rhat >= b)
break;
p = p - vn0;
}
q1 = qhat & 0xFFFFFFFF;
#else // ORIGINAL
q1 = un64 / vn1;
8 rhat = un64 - q1 * vn1;
9 while (q1 >= b || q1 * vn0 > b * rhat + un1) {
q1 = q1 - 1;
rhat = rhat + vn1;
if (rhat >= b)
break;
}
#endif
// Multiply and subtract
un21 = un64 * b + un1 - q1 * v;
// Compute the second quotient digit
#ifdef OPTIMIZE
qhat = un21 / vn1;
rhat = un21 % vn1;
p = qhat * vn0;
while (qhat >= b || p > b * rhat + un0) {
qhat = qhat - 1;
rhat = rhat + vn1;
if (rhat >= b)
break;
p = p - vn0;
}
q0 = qhat & 0xFFFFFFFF;
#else // ORIGINAL
q0 = un21 / vn1;
8 rhat = un21 - q0 * vn1;
9 while (q0 >= b || q0 * vn0 > b * rhat + un0) {
q0 = q0 - 1;
rhat = rhat + vn1;
if (rhat >= b)
break;
}
#endif
// If remainder is wanted, return it
if (r != NULL)
*r = (un21 * b + un0 - q0 * v) >> s;
return q1 * b + q0;
}
…- The number base is 232, i.e. 32 bits, not 16 bits, as stated in the comment.
- The
normalised
(least significant) digits of dividend and divisor as well as the final digits of the quotient fit in 32 bits, i.e. the variablesun1,un0,vn1,vn0,q1andq0can and should of course be declared asuint32_t, not asuint64_t. - The proper name for the variable with the two most significant digits of the dividend is
un32, notun64. - The variables
unand6432un10are superfluous, the dividendu1andu0can and should of course be normalised in place, like the divisorv. - The term
& (-s >> 31)is superfluous: in the original code it is used to avoid the conditional expressionif (s > 0) … else …and evaluates to either& 0or& -1(alias& 0xFFFFFFFFor& ~0) there; here it always evaluates to& -1(alias& 0xFFFFFFFFFFFFFFFFor& ~0LL). - The term
| u0produces wrong results and must be removed. - The variables
un1,un0,vn1andvn0are superfluous, they can and of course should be replaced by the expressions(uint32_t) (un10 >> 32),(uint32_t) (un10 & 0xFFFFFFFF),(uint32_t) (v >> 32)and(uint32_t) (v & 0xFFFFFFFF)respectively. - The remainder
rhatcan and should of course be computed asunand6432 % vn1un21 % vn1respectively, not asunand6432 - q1 * vn1un21 - q0 * vn1using expensive 64×64-bit multiplications: both hardware division instructions and software division routines typically return quotient and remainder together. - The repeated (expensive) computation of the 64×64-bit products
q1 * vn0andq0 * vn0inside thewhile …loops should be moved outside the loops.
Poor implementation in Free Pascal’s Run-Time Library
… flt_core.inc from Free Pascal’s Run-Time Library…
…
(*-------------------------------------------------------
| u128_div_u64_to_u64 [local]
|
| Divides unsigned 128-bit integer by unsigned 64-bit integer.
| Returns 64-bit quotient and reminder.
|
| This routine is used here only for splitting specially prepared unsigned
| 128-bit integer into two 64-bit ones before converting it to ASCII.
|
*-------------------------------------------------------*)
function u128_div_u64_to_u64( const xh, xl: qword; const y: qword; out quotient, reminder: qword ): boolean;
var
b, // Number base
1 v, // Norm. divisor
2 un1, un0, // Norm. dividend LSD's
2 vn1, vn0, // Norm. divisor digits
q1, q0, // Quotient digits
3 un64, un21, un10, // Dividend digit pairs
rhat: qword; // A remainder
s: integer; // Shift amount for norm
begin
// Overflow check
if ( xh >= y ) then
begin
u128_div_u64_to_u64 := false;
exit;
end;
// Count leading zeros
s := 63 - BSRqword( y ); // 0 <= s <= 63
// Normalize divisor
1 v := y shl s;
// Break divisor up into two 32-bit digits
4 vn1 := hi(v);
4 vn0 := lo(v);
// Shift dividend left
1 un64 := xh shl s;
if ( s > 0 ) then
1 un64 := un64 or ( xl shr ( 64 - s ) );
1 un10 := xl shl s;
// Break right half of dividend into two digits
4 un1 := hi(un10);
4 un0 := lo(un10);
// Compute the first quotient digit, q1
3 q1 := un64 div vn1;
5 rhat := un64 - q1 * vn1;
b := qword(1) shl 32; // Number base
6 while ( q1 >= b )
7 or ( q1 * vn0 > b * rhat + un1 ) do
begin
dec( q1 );
inc( rhat, vn1 );
6 if rhat >= b then
break;
end;
// Multiply and subtract
3 un21 := un64 * b + un1 - q1 * v;
// Compute the second quotient digit, q0
q0 := un21 div vn1;
5 rhat := un21 - q0 * vn1;
6 while ( q0 >= b )
7 or ( q0 * vn0 > b * rhat + un0 ) do
begin
dec( q0 );
inc( rhat, vn1 );
6 if ( rhat >= b ) then
break;
end;
// Result
8 reminder := ( un21 * b + un0 - q0 * v ) shr s;
8 quotient := q1 * b + q0;
u128_div_u64_to_u64 := true;
end;
…- The variables
v,unand6432un10are superfluous, the divisoryas well as the dividendxhandxlcan and should of course be normalised in place. - The
normalised
(least significant) digits of dividend and divisor fit in 32 bits, i.e. the variablesun1,un0,vn1andvn0can and should of course be declared asdword, not asqword. - The proper name for the variable with the two most significant digits of the dividend is
un32, notun64. - The variables
un1,un0,vn1andvn0are superfluous, they can and of course should be replaced by the expressionshi(un10),lo(un10),hi(v)andlow(v)respectively. - The remainder
rhatcan and should of course be computed asunand6432 mod vn1un21 mod vn1respectively, not asunand6432 - q1 * vn1un21 - q0 * vn1using expensive 64×64-bit multiplications: both hardware division instructions and software division routines typically return quotient and remainder together. - The comparisions
q1 >= b,q0 >= bandrhat >= bcan and should of course be replaced byhi(q1) <> 0,hi(q0) <> 0andhi(rhat) <> 0respectively. - The repeated (expensive) computation of the 64×64-bit products
q1 * vn0andq0 * vn0inside thewhile …loops should be moved outside the loops. - The final quotient digits fit in 32 bits, i.e. the variables
q1andq0can and should of course be replaced wherever possible bylo(q1)andlo(q0)respectively to avoid expensive 64×64-bit multiplications.
Poor implementation in Google’s Go Programming Language
… bits.go …Note: the highlighted parts show the optimisations and simplifications missed by the author(s) of the Div64 function!
// Copyright 2017 The Go Authors. All rights reserved.
…
// Div64 returns the quotient and remainder of (hi, lo) divided by y:
// quo = (hi, lo)/y, rem = (hi, lo)%y with the dividend bits' upper
// half in parameter hi and the lower half in parameter lo.
// Div64 panics for y == 0 (division by zero) or y <= hi (quotient overflow).
func Div64(hi, lo, y uint64) (quo, rem uint64) {
const (
two32 = 1 << 32
mask32 = two32 - 1
)
if y == 0 {
panic(divideError)
}
if y <= hi {
panic(overflowError)
}
s := uint(LeadingZeros64(y))
y <<= s
1 yn1 := uint32(y >> 32)
1 yn0 := uint32(y & mask32)
un32 := hi<<s | lo>>(64-s)
un10 := lo << s
1 un1 := uint32(un10 >> 32)
1 un0 := uint32(un10 & mask32)
q1 := un32 / yn1
2 rhat := un32 - q1*yn1
3 for q1 >= two32 || uint32(q1)*yn0 > two32*rhat+un1 {
q1--
rhat += yn1
if rhat >= two32 {
break
}
}
3 un21 := un32*two32 + un1 - uint32(q1)*y
q0 := un21 / yn1
2 rhat = un21 - q0*yn1
3 for q0 >= two32 || uint32(q0)*yn0 > two32*rhat+un0 {
q0--
rhat += yn1
if rhat >= two32 {
break
}
}
3 return q1*two32 + q0, (un21*two32 + un0 - uint32(q0)*y) >> s
}
…- The
normalised
(least significant) digits of dividend and divisor fit in 32 bits, i.e. the variablesyn1,yn0,un1andun0can and should of course be declared asuint32, not asuint64. - The remainder
rhatcan and should of course be computed asun32 % yn1andun21 % yn1respectively, not asun32 - q1*yn1andun21 - q0*yn1using expensive 64×64-bit multiplications: both hardware division instructions and software division routines typically return quotient and remainder together. - The final digits of the quotient fit in 32 bits, i.e. the variables
q1andq0can and should of course be cast touint32wherever necessary to avoid expensive 64×64-bit multiplications.
Implementation in Google’s V8 JavaScript engine
… bigint.cc from Google’s V8 JavaScript engine.Note: the type digit_t is the same as uintptr_t.
// Copyright 2017 the V8 project authors. All rights reserved.
…
// Returns the quotient.
// quotient = (high << kDigitBits + low - remainder) / divisor
BigInt::digit_t MutableBigInt::digit_div(digit_t high, digit_t low,
digit_t divisor, digit_t* remainder) {
DCHECK(high < divisor);
#if V8_TARGET_ARCH_X64 && (__GNUC__ || __clang__)
…
#elif V8_TARGET_ARCH_IA32 && (__GNUC__ || __clang__)
…
#else
static const digit_t kHalfDigitBase = 1ull << kHalfDigitBits;
// Adapted from Warren, Hacker's Delight, p. 152.
int s = base::bits::CountLeadingZeros(divisor);
DCHECK_NE(s, kDigitBits); // {divisor} is not 0.
divisor <<= s;
digit_t vn1 = divisor >> kHalfDigitBits;
digit_t vn0 = divisor & kHalfDigitMask;
// {s} can be 0. {low >> kDigitBits} would be undefined behavior, so
// we mask the shift amount with {kShiftMask}, and the result with
// {s_zero_mask} which is 0 if s == 0 and all 1-bits otherwise.
STATIC_ASSERT(sizeof(intptr_t) == sizeof(digit_t));
const int kShiftMask = kDigitBits - 1;
digit_t s_zero_mask =
static_cast<digit_t>(static_cast<intptr_t>(-s) >> (kDigitBits - 1));
digit_t un32 =
(high << s) | ((low >> ((kDigitBits - s) & kShiftMask)) & s_zero_mask);
digit_t un10 = low << s;
digit_t un1 = un10 >> kHalfDigitBits;
digit_t un0 = un10 & kHalfDigitMask;
digit_t q1 = un32 / vn1;
digit_t rhat = un32 - q1 * vn1;
while (q1 >= kHalfDigitBase || q1 * vn0 > rhat * kHalfDigitBase + un1) {
q1--;
rhat += vn1;
if (rhat >= kHalfDigitBase) break;
}
digit_t un21 = un32 * kHalfDigitBase + un1 - q1 * divisor;
digit_t q0 = un21 / vn1;
rhat = un21 - q0 * vn1;
while (q0 >= kHalfDigitBase || q0 * vn0 > rhat * kHalfDigitBase + un0) {
q0--;
rhat += vn1;
if (rhat >= kHalfDigitBase) break;
}
*remainder = (un21 * kHalfDigitBase + un0 - q0 * divisor) >> s;
return q1 * kHalfDigitBase + q0;
#endif
}
…Proper (and optimised) implementation …
… in ANSI C
This further simplified and optimised implementation ofAlgorithm Dfor unsigned 128÷64-bit division on 32-bit machines is based on a 64÷32-bit division returning a 64-bit quotient and a 32-bit remainder, trivially implemented per
long(alias
schoolbook) division using a
narrowing64÷32-bit division returning a 32-bit quotient and a 32-bit remainder.
// Copyleft © 2011-2021, Stefan Kanthak <stefan.kanthak@nexgo.de>
// Divide a 128-bit integer dividend, supplied as a pair of 64-bit
// integers in u0 and u1, by a 64-bit integer divisor, supplied in v;
// return the 64-bit quotient and optionally the 64-bit remainder in *r.
unsigned long long divllu(unsigned long long u0,
unsigned long long u1,
unsigned long long v,
unsigned long long *r) {
unsigned long long qhat; // A quotient.
unsigned long long rhat; // A remainder.
unsigned long long uhat; // A dividend digit pair.
unsigned q0, q1; // Quotient digits.
unsigned s; // Shift amount for norm.
if (u1 >= v) { // If overflow, set rem.
if (r != NULL) // to an impossible value,
*r = ~0ULL; // and return the largest
return ~0ULL; // possible quotient.
}
s = __builtin_clzll(v); // 0 <= s <= 63.
if (s != 0) {
v <<= s; // Normalize divisor.
u1 <<= s; // Shift dividend left.
u1 |= u0 >> (64 - s);
u0 <<= s;
}
// Compute high quotient digit.
qhat = u1 / (unsigned) (v >> 32);
rhat = u1 % (unsigned) (v >> 32);
while ((unsigned) (qhat >> 32) != 0U ||
// Both qhat and rhat are less 2**32 here!
(unsigned long long) (unsigned) (qhat & ~0U) * (unsigned) (v & ~0U) >
((rhat << 32) | (unsigned) (u0 >> 32))) {
qhat -= 1;
rhat += (unsigned) (v >> 32);
if ((unsigned) (rhat >> 32) != 0U) break;
}
q1 = (unsigned) (qhat & ~0U);
// Multiply and subtract.
uhat = ((u1 << 32) | (unsigned) (u0 >> 32)) - q1 * v;
// Compute low quotient digit.
qhat = uhat / (unsigned) (v >> 32);
rhat = uhat % (unsigned) (v >> 32);
while ((unsigned) (qhat >> 32) != 0U ||
// Both qhat and rhat are less 2**32 here!
(unsigned long long) (unsigned) (qhat & ~0U) * (unsigned) (v & ~0U) >
((rhat << 32) | (unsigned) (u0 & ~0U))) {
qhat -= 1;
rhat += (unsigned) (v >> 32);
if ((unsigned) (rhat >> 32) != 0U) break;
}
q0 = (unsigned) (qhat & ~0U);
if (r != NULL) // If remainder is wanted, return it.
*r = (((uhat << 32) | (unsigned) (u0 & ~0U)) - q0 * v) >> s;
return ((unsigned long long) q1 << 32) | q0;
}… in i386 Assembler
Finally the implementation of this optimised simplified variant in assembly language, with yet another optimisation: it performs along(alias
schoolbook) division for divisors less than 232 and dividends less than 296.
; Copyright © 2011-2021, Stefan Kanthak <stefan.kanthak@nexgo.de>
.386
.model flat, C
.code
divllu proc public
push ebx
push ebp
push edi
push esi
…
mov esi, [esp+40]
mov edi, [esp+36] ; esi:edi = divisor
bsr ecx, esi ; ecx = index of leading '1' bit in high dword of divisor
jnz NORMALIZE ; high dword of divisor (and high dword of dividend) = 0?
; "long" division (dividend < 2**96, divisor < 2**32)
DIVISION:
mov edx, [esp+28]
mov eax, [esp+24] ; edx:eax = "inner" qword of dividend
div edi ; eax = high dword of quotient,
; edx = high dword of dividend'
mov ebx, eax ; ebx = high dword of quotient
mov eax, [esp+20] ; edx:eax = low qword of dividend'
div edi ; eax = low dword of quotient,
; edx = (low dword) of remainder
or esi, [esp+44] ; esi = address of remainder
jz @F ; address of remainder = 0?
xor ecx, ecx ; ecx:edx = remainder
mov [esi], edx
mov [esi+4], ecx ; store remainder
@@:
mov edx, ebx ; edx:eax = quotient
pop esi
pop edi
pop ebp
pop ebx
ret
NORMALIZE:
mov edx, [esp+32]
mov eax, [esp+28] ; edx:eax = high qword of dividend
xor ecx, 31 ; ecx = number of leading '0' bits in (high dword of) divisor
jz @F ; number of leading '0' bits = 0?
shld esi, edi, cl
shl edi, cl ; esi:edi = divisor'
mov [esp+40], esi
mov [esp+36], edi ; save normalised divisor'
mov ebx, [esp+24]
mov ebp, [esp+20] ; ebx:ebp = low qword of dividend
shld edx, eax, cl
shld eax, ebx, cl
shld ebx, ebp, cl
shl ebp, cl ; edx:eax:ebx:ebp = dividend'
mov [esp+32], edx
mov [esp+28], eax
mov [esp+24], ebx
mov [esp+20], ebp ; save normalised dividend'
@@:
push ecx ; save number of leading '0' bits in divisor
DIVISION_1:
cmp esi, edx
jna OVERFLOW_1 ; overflow with normal division?
; "normal" division
NORMAL_1:
div esi ; eax = (low dword of) quotient',
; edx = (low dword of) remainder'
mov ebp, edx ; ebp = (low dword of) remainder'
mov ecx, eax
xor ebx, ebx ; ebx:ecx = quotient'
jmp CHECK_1
; "long" division
OVERFLOW_1:
mov ecx, eax
mov eax, edx
xor edx, edx
div esi ; eax = high dword of quotient',
; edx = high dword of remainder'
mov ebx, eax ; ebx = high dword of quotient'
mov eax, ecx
div esi ; eax = low dword of quotient',
; edx = (low dword of) remainder'
mov ebp, edx ; ebp = (low dword of) remainder'
mov ecx, eax ; ebx:ecx = quotient'
ADJUST_1:
add ecx, -1
adc ebx, -1 ; ebx:ecx = quotient' - 1
add ebp, esi ; ebp = (low dword of) remainder'
; + high dword of divisor'
jc BREAK_1 ; remainder' >= 2**32?
AGAIN_1:
test ebx, ebx
jnz ADJUST_1 ; quotient' >= 2**32?
CHECK_1:
mov eax, edi ; eax = low dword of divisor'
mul ecx ; edx:eax = low dword of divisor'
; * low dword of quotient'
ifdef JMPLESS
cmp [esp+28], eax
mov eax, ebp
sbb eax, edx
jb ADJUST_1
else
cmp edx, ebp
jb BREAK_1
ja ADJUST_1
cmp eax, [esp+28]
ja ADJUST_1
endif
BREAK_1:
push ecx ; save (low dword of) quotient'
mov eax, edi ; eax = low dword of divisor'
mul ecx ; edx:eax = low dword of divisor'
; * (low dword of) quotient'
imul ecx, esi ; ecx = (low dword of) quotient'
; * high dword of divisor'
add ecx, edx ; ecx:eax = divisor'
; * (low dword of) quotient'
mov ebx, eax ; ecx:ebx = divisor'
; * (low dword of) quotient'
mov eax, [esp+32]
mov edx, [esp+36] ; edx:eax = "inner" qword of dividend'
sub eax, ebx
sbb edx, ecx ; edx:eax = "inner" qword of dividend"
; = intermediate (normalised) remainder
push eax ; save low dword of "inner" qword of dividend"
DIVISION_2:
cmp esi, edx
jna OVERFLOW_2 ; overflow with normal division?
; "normal" division
NORMAL_2:
div esi ; eax = (low dword of) quotient",
; edx = (low dword of) remainder"
mov ebp, edx ; ebp = (low dword of) remainder"
mov ecx, eax
xor ebx, ebx ; ebx:ecx = quotient"
jmp CHECK_2
; "long" division
OVERFLOW_2:
mov ecx, eax
mov eax, edx
xor edx, edx
div esi ; eax = high dword of quotient",
; edx = high dword of remainder"
mov ebx, eax ; ebx = high dword of quotient"
mov eax, ecx
div esi ; eax = low dword of quotient",
; edx = (low dword of) remainder"
mov ebp, edx ; ebp = (low dword of) remainder"
mov ecx, eax ; ebx:ecx = quotient"
ADJUST_2:
add ecx, -1
adc ebx, -1 ; ebx:ecx = quotient" - 1
add ebp, esi ; ebp = (low dword of) remainder"
; + high dword of divisor'
jc BREAK_2 ; remainder" >= 2**32?
AGAIN_2:
test ebx, ebx
jnz ADJUST_2 ; quotient" >= 2**32?
CHECK_2:
mov eax, edi ; eax = low dword of divisor'
mul ecx ; edx:eax = low dword of divisor'
; * low dword of quotient"
ifdef JMPLESS
cmp [esp+32], eax
mov eax, ebp
sbb eax, edx
jb ADJUST_2
else
cmp edx, ebp
jb BREAK_2
ja ADJUST_2
cmp eax, [esp+32]
ja ADJUST_2
endif
BREAK_2:
pop ebx ; ebx = low dword of "inner" qword of dividend"
push ecx ; save (low dword of) quotient"
mov ebp, [esp+56] ; ebp = address of remainder
test ebp, ebp
jz QUOTIENT ; address of remainder = 0?
REMAINDER:
mov eax, edi ; eax = low dword of divisor'
mul ecx ; edx:eax = low dword of divisor'
; * (low dword of) quotient"
imul ecx, esi ; ecx = (low dword of) quotient"
; * high dword of divisor'
add edx, ecx ; edx:eax = (low dword of) quotient"
; * divisor'
mov edi, [esp+32]
mov esi, ebx
sub edi, eax
sbb esi, edx ; esi:edi = (normalised) remainder
mov ecx, [esp+8] ; ecx = number of leading '0' bits in divisor
; = shift count
if 0
jecxz @F ; shift count = 0?
else
test ecx, ecx
jz @F ; shift count = 0?
endif
shrd edi, esi, cl
shr esi, cl ; esi:edi = remainder
@@:
mov [ebp], edi
mov [ebp+4], esi ; store remainder
QUOTIENT:
pop eax
pop edx ; edx:eax = quotient
pop ecx
pop esi
pop edi
pop ebp
pop ebx
ret
divllu endp
endContact
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