Fastest way to calculate a 128-bit integer modulo a 64-bit integer - c

Question

I have a 128-bit unsigned integer A and a 64-bit unsigned integer B. What's the fastest way to calculate A % B - that is the (64-bit) remainder from dividing A by B?
I'm looking to do this in either C or assembly language, but I need to target the 32-bit x86 platform. This unfortunately means that I cannot take advantage of compiler support for 128-bit integers, nor of the x64 architecture's ability to perform the required operation in a single instruction.
Edit:
Thank you for the answers so far. However, it appears to me that the suggested algorithms would be quite slow - wouldn't the fastest way to perform a 128-bit by 64-bit division be to leverage the processor's native support for 64-bit by 32-bit division? Does anyone know if there is a way to perform the larger division in terms of a few smaller divisions?
Re: How often does B change?
Primarily I'm interested in a general solution - what calculation would you perform if A and B are likely to be different every time?
However, a second possible situation is that B does not vary as often as A - there may be as many as 200 As to divide by each B. How would your answer differ in this case?

Answer 1

You can use the division version of Russian Peasant Multiplication.
To find the remainder, execute (in pseudo-code):
X = B;
while (X <= A/2)
{
X <<= 1;
}
while (A >= B)
{
if (A >= X)
A -= X;
X >>= 1;
}
The modulus is left in A.
You'll need to implement the shifts, comparisons and subtractions to operate on values made up of a pair of 64 bit numbers, but that's fairly trivial (likely you should implement the left-shift-by-1 as X + X).
This will loop at most 255 times (with a 128 bit A). Of course you need to do a pre-check for a zero divisor.

Answer 2

Perhaps you're looking for a finished program, but the basic algorithms for multi-precision arithmetic can be found in Knuth's Art of Computer Programming, Volume 2. You can find the division algorithm described online here. The algorithms deal with arbitrary multi-precision arithmetic, and so are more general than you need, but you should be able to simplify them for 128 bit arithmetic done on 64- or 32-bit digits. Be prepared for a reasonable amount of work (a) understanding the algorithm, and (b) converting it to C or assembler.
You might also want to check out Hacker's Delight, which is full of very clever assembler and other low-level hackery, including some multi-precision arithmetic.

Answer 3

If your B is small enough for the uint64_t + operation to not wrap:
Given A = AH*2^64 + AL:
A % B == (((AH % B) * (2^64 % B)) + (AL % B)) % B
== (((AH % B) * ((2^64 - B) % B)) + (AL % B)) % B
If your compiler supports 64-bit integers, then this is probably the easiest way to go.
MSVC's implementation of a 64-bit modulo on 32-bit x86 is some hairy loop filled assembly (VC\crt\src\intel\llrem.asm for the brave), so I'd personally go with that.

Answer 4

This is almost untested partly speed modificated Mod128by64 'Russian peasant' algorithm function. Unfortunately I'm a Delphi user so this function works under Delphi. :) But the assembler is almost the same so...
function Mod128by64(Dividend: PUInt128; Divisor: PUInt64): UInt64;
//In : eax = #Dividend
// : edx = #Divisor
//Out: eax:edx as Remainder
asm
//Registers inside rutine
//Divisor = edx:ebp
//Dividend = bh:ebx:edx //We need 64 bits + 1 bit in bh
//Result = esi:edi
//ecx = Loop counter and Dividend index
push ebx //Store registers to stack
push esi
push edi
push ebp
mov ebp, [edx] //Divisor = edx:ebp
mov edx, [edx + 4]
mov ecx, ebp //Div by 0 test
or ecx, edx
jz #DivByZero
xor edi, edi //Clear result
xor esi, esi
//Start of 64 bit division Loop
mov ecx, 15 //Load byte loop shift counter and Dividend index
#SkipShift8Bits: //Small Dividend numbers shift optimisation
cmp [eax + ecx], ch //Zero test
jnz #EndSkipShiftDividend
loop #SkipShift8Bits //Skip 8 bit loop
#EndSkipShiftDividend:
test edx, $FF000000 //Huge Divisor Numbers Shift Optimisation
jz #Shift8Bits //This Divisor is > $00FFFFFF:FFFFFFFF
mov ecx, 8 //Load byte shift counter
mov esi, [eax + 12] //Do fast 56 bit (7 bytes) shift...
shr esi, cl //esi = $00XXXXXX
mov edi, [eax + 9] //Load for one byte right shifted 32 bit value
#Shift8Bits:
mov bl, [eax + ecx] //Load 8 bits of Dividend
//Here we can unrole partial loop 8 bit division to increase execution speed...
mov ch, 8 //Set partial byte counter value
#Do65BitsShift:
shl bl, 1 //Shift dividend left for one bit
rcl edi, 1
rcl esi, 1
setc bh //Save 65th bit
sub edi, ebp //Compare dividend and divisor
sbb esi, edx //Subtract the divisor
sbb bh, 0 //Use 65th bit in bh
jnc #NoCarryAtCmp //Test...
add edi, ebp //Return privius dividend state
adc esi, edx
#NoCarryAtCmp:
dec ch //Decrement counter
jnz #Do65BitsShift
//End of 8 bit (byte) partial division loop
dec cl //Decrement byte loop shift counter
jns #Shift8Bits //Last jump at cl = 0!!!
//End of 64 bit division loop
mov eax, edi //Load result to eax:edx
mov edx, esi
#RestoreRegisters:
pop ebp //Restore Registers
pop edi
pop esi
pop ebx
ret
#DivByZero:
xor eax, eax //Here you can raise Div by 0 exception, now function only return 0.
xor edx, edx
jmp #RestoreRegisters
end;
At least one more speed optimisation is possible! After 'Huge Divisor Numbers Shift Optimisation' we can test divisors high bit, if it is 0 we do not need to use extra bh register as 65th bit to store in it. So unrolled part of loop can look like:
shl bl,1 //Shift dividend left for one bit
rcl edi,1
rcl esi,1
sub edi, ebp //Compare dividend and divisor
sbb esi, edx //Subtract the divisor
jnc #NoCarryAtCmpX
add edi, ebp //Return privius dividend state
adc esi, edx
#NoCarryAtCmpX:

Answer 5

I know the question specified 32-bit code, but the answer for 64-bit may be useful or interesting to others.
And yes, 64b/32b => 32b division does make a useful building-block for 128b % 64b => 64b. libgcc's __umoddi3 (source linked below) gives an idea of how to do that sort of thing, but it only implements 2N % 2N => 2N on top of a 2N / N => N division, not 4N % 2N => 2N.
Wider multi-precision libraries are available, e.g. https://gmplib.org/manual/Integer-Division.html#Integer-Division.
GNU C on 64-bit machines does provide an __int128 type, and libgcc functions to multiply and divide as efficiently as possible on the target architecture.
x86-64's div r/m64 instruction does 128b/64b => 64b division (also producing remainder as a second output), but it faults if the quotient overflows. So you can't directly use it if A/B > 2^64-1, but you can get gcc to use it for you (or even inline the same code that libgcc uses).
This compiles (Godbolt compiler explorer) to one or two div instructions (which happen inside a libgcc function call). If there was a faster way, libgcc would probably use that instead.
#include <stdint.h>
uint64_t AmodB(unsigned __int128 A, uint64_t B) {
return A % B;
}
The __umodti3 function it calls calculates a full 128b/128b modulo, but the implementation of that function does check for the special case where the divisor's high half is 0, as you can see in the libgcc source. (libgcc builds the si/di/ti version of the function from that code, as appropriate for the target architecture. udiv_qrnnd is an inline asm macro that does unsigned 2N/N => N division for the target architecture.
For x86-64 (and other architectures with a hardware divide instruction), the fast-path (when high_half(A) < B; guaranteeing div won't fault) is just two not-taken branches, some fluff for out-of-order CPUs to chew through, and a single div r64 instruction, which takes about 50-100 cycles1 on modern x86 CPUs, according to Agner Fog's insn tables. Some other work can be happening in parallel with div, but the integer divide unit is not very pipelined and div decodes to a lot of uops (unlike FP division).
The fallback path still only uses two 64-bit div instructions for the case where B is only 64-bit, but A/B doesn't fit in 64 bits so A/B directly would fault.
Note that libgcc's __umodti3 just inlines __udivmoddi4 into a wrapper that only returns the remainder.
Footnote 1: 32-bit div is over 2x faster on Intel CPUs. On AMD CPUs, performance only depends on the size of the actual input values, even if they're small values in a 64-bit register. If small values are common, it might be worth benchmarking a branch to a simple 32-bit division version before doing 64-bit or 128-bit division.
For repeated modulo by the same B
It might be worth considering calculating a fixed-point multiplicative inverse for B, if one exists. For example, with compile-time constants, gcc does the optimization for types narrower than 128b.
uint64_t modulo_by_constant64(uint64_t A) { return A % 0x12345678ABULL; }
movabs rdx, -2233785418547900415
mov rax, rdi
mul rdx
mov rax, rdx # wasted instruction, could have kept using RDX.
movabs rdx, 78187493547
shr rax, 36 # division result
imul rax, rdx # multiply and subtract to get the modulo
sub rdi, rax
mov rax, rdi
ret
x86's mul r64 instruction does 64b*64b => 128b (rdx:rax) multiplication, and can be used as a building block to construct a 128b * 128b => 256b multiply to implement the same algorithm. Since we only need the high half of the full 256b result, that saves a few multiplies.
Modern Intel CPUs have very high performance mul: 3c latency, one per clock throughput. However, the exact combination of shifts and adds required varies with the constant, so the general case of calculating a multiplicative inverse at run-time isn't quite as efficient each time its used as a JIT-compiled or statically-compiled version (even on top of the pre-computation overhead).
IDK where the break-even point would be. For JIT-compiling, it will be higher than ~200 reuses, unless you cache generated code for commonly-used B values. For the "normal" way, it might possibly be in the range of 200 reuses, but IDK how expensive it would be to find a modular multiplicative inverse for 128-bit / 64-bit division.
libdivide can do this for you, but only for 32 and 64-bit types. Still, it's probably a good starting point.

Answer 6

I have made both version of Mod128by64 'Russian peasant' division function: classic and speed optimised. Speed optimised can do on my 3Ghz PC more than 1000.000 random calculations per second and is more than three times faster than classic function.
If we compare the execution time of calculating 128 by 64 and calculating 64 by 64 bit modulo than this function is only about 50% slower.
Classic Russian peasant:
function Mod128by64Clasic(Dividend: PUInt128; Divisor: PUInt64): UInt64;
//In : eax = #Dividend
// : edx = #Divisor
//Out: eax:edx as Remainder
asm
//Registers inside rutine
//edx:ebp = Divisor
//ecx = Loop counter
//Result = esi:edi
push ebx //Store registers to stack
push esi
push edi
push ebp
mov ebp, [edx] //Load divisor to edx:ebp
mov edx, [edx + 4]
mov ecx, ebp //Div by 0 test
or ecx, edx
jz #DivByZero
push [eax] //Store Divisor to the stack
push [eax + 4]
push [eax + 8]
push [eax + 12]
xor edi, edi //Clear result
xor esi, esi
mov ecx, 128 //Load shift counter
#Do128BitsShift:
shl [esp + 12], 1 //Shift dividend from stack left for one bit
rcl [esp + 8], 1
rcl [esp + 4], 1
rcl [esp], 1
rcl edi, 1
rcl esi, 1
setc bh //Save 65th bit
sub edi, ebp //Compare dividend and divisor
sbb esi, edx //Subtract the divisor
sbb bh, 0 //Use 65th bit in bh
jnc #NoCarryAtCmp //Test...
add edi, ebp //Return privius dividend state
adc esi, edx
#NoCarryAtCmp:
loop #Do128BitsShift
//End of 128 bit division loop
mov eax, edi //Load result to eax:edx
mov edx, esi
#RestoreRegisters:
lea esp, esp + 16 //Restore Divisors space on stack
pop ebp //Restore Registers
pop edi
pop esi
pop ebx
ret
#DivByZero:
xor eax, eax //Here you can raise Div by 0 exception, now function only return 0.
xor edx, edx
jmp #RestoreRegisters
end;
Speed optimised Russian peasant:
function Mod128by64Oprimized(Dividend: PUInt128; Divisor: PUInt64): UInt64;
//In : eax = #Dividend
// : edx = #Divisor
//Out: eax:edx as Remainder
asm
//Registers inside rutine
//Divisor = edx:ebp
//Dividend = ebx:edx //We need 64 bits
//Result = esi:edi
//ecx = Loop counter and Dividend index
push ebx //Store registers to stack
push esi
push edi
push ebp
mov ebp, [edx] //Divisor = edx:ebp
mov edx, [edx + 4]
mov ecx, ebp //Div by 0 test
or ecx, edx
jz #DivByZero
xor edi, edi //Clear result
xor esi, esi
//Start of 64 bit division Loop
mov ecx, 15 //Load byte loop shift counter and Dividend index
#SkipShift8Bits: //Small Dividend numbers shift optimisation
cmp [eax + ecx], ch //Zero test
jnz #EndSkipShiftDividend
loop #SkipShift8Bits //Skip Compute 8 Bits unroled loop ?
#EndSkipShiftDividend:
test edx, $FF000000 //Huge Divisor Numbers Shift Optimisation
jz #Shift8Bits //This Divisor is > $00FFFFFF:FFFFFFFF
mov ecx, 8 //Load byte shift counter
mov esi, [eax + 12] //Do fast 56 bit (7 bytes) shift...
shr esi, cl //esi = $00XXXXXX
mov edi, [eax + 9] //Load for one byte right shifted 32 bit value
#Shift8Bits:
mov bl, [eax + ecx] //Load 8 bit part of Dividend
//Compute 8 Bits unroled loop
shl bl, 1 //Shift dividend left for one bit
rcl edi, 1
rcl esi, 1
jc #DividentAbove0 //dividend hi bit set?
cmp esi, edx //dividend hi part larger?
jb #DividentBelow0
ja #DividentAbove0
cmp edi, ebp //dividend lo part larger?
jb #DividentBelow0
#DividentAbove0:
sub edi, ebp //Return privius dividend state
sbb esi, edx
#DividentBelow0:
shl bl, 1 //Shift dividend left for one bit
rcl edi, 1
rcl esi, 1
jc #DividentAbove1 //dividend hi bit set?
cmp esi, edx //dividend hi part larger?
jb #DividentBelow1
ja #DividentAbove1
cmp edi, ebp //dividend lo part larger?
jb #DividentBelow1
#DividentAbove1:
sub edi, ebp //Return privius dividend state
sbb esi, edx
#DividentBelow1:
shl bl, 1 //Shift dividend left for one bit
rcl edi, 1
rcl esi, 1
jc #DividentAbove2 //dividend hi bit set?
cmp esi, edx //dividend hi part larger?
jb #DividentBelow2
ja #DividentAbove2
cmp edi, ebp //dividend lo part larger?
jb #DividentBelow2
#DividentAbove2:
sub edi, ebp //Return privius dividend state
sbb esi, edx
#DividentBelow2:
shl bl, 1 //Shift dividend left for one bit
rcl edi, 1
rcl esi, 1
jc #DividentAbove3 //dividend hi bit set?
cmp esi, edx //dividend hi part larger?
jb #DividentBelow3
ja #DividentAbove3
cmp edi, ebp //dividend lo part larger?
jb #DividentBelow3
#DividentAbove3:
sub edi, ebp //Return privius dividend state
sbb esi, edx
#DividentBelow3:
shl bl, 1 //Shift dividend left for one bit
rcl edi, 1
rcl esi, 1
jc #DividentAbove4 //dividend hi bit set?
cmp esi, edx //dividend hi part larger?
jb #DividentBelow4
ja #DividentAbove4
cmp edi, ebp //dividend lo part larger?
jb #DividentBelow4
#DividentAbove4:
sub edi, ebp //Return privius dividend state
sbb esi, edx
#DividentBelow4:
shl bl, 1 //Shift dividend left for one bit
rcl edi, 1
rcl esi, 1
jc #DividentAbove5 //dividend hi bit set?
cmp esi, edx //dividend hi part larger?
jb #DividentBelow5
ja #DividentAbove5
cmp edi, ebp //dividend lo part larger?
jb #DividentBelow5
#DividentAbove5:
sub edi, ebp //Return privius dividend state
sbb esi, edx
#DividentBelow5:
shl bl, 1 //Shift dividend left for one bit
rcl edi, 1
rcl esi, 1
jc #DividentAbove6 //dividend hi bit set?
cmp esi, edx //dividend hi part larger?
jb #DividentBelow6
ja #DividentAbove6
cmp edi, ebp //dividend lo part larger?
jb #DividentBelow6
#DividentAbove6:
sub edi, ebp //Return privius dividend state
sbb esi, edx
#DividentBelow6:
shl bl, 1 //Shift dividend left for one bit
rcl edi, 1
rcl esi, 1
jc #DividentAbove7 //dividend hi bit set?
cmp esi, edx //dividend hi part larger?
jb #DividentBelow7
ja #DividentAbove7
cmp edi, ebp //dividend lo part larger?
jb #DividentBelow7
#DividentAbove7:
sub edi, ebp //Return privius dividend state
sbb esi, edx
#DividentBelow7:
//End of Compute 8 Bits (unroled loop)
dec cl //Decrement byte loop shift counter
jns #Shift8Bits //Last jump at cl = 0!!!
//End of division loop
mov eax, edi //Load result to eax:edx
mov edx, esi
#RestoreRegisters:
pop ebp //Restore Registers
pop edi
pop esi
pop ebx
ret
#DivByZero:
xor eax, eax //Here you can raise Div by 0 exception, now function only return 0.
xor edx, edx
jmp #RestoreRegisters
end;

Answer 7

I'd like to share a few thoughts.
It's not as simple as MSN proposes I'm afraid.
In the expression:
(((AH % B) * ((2^64 - B) % B)) + (AL % B)) % B
both multiplication and addition may overflow. I think one could take it into account and still use the general concept with some modifications, but something tells me it's going to get really scary.
I was curious how 64 bit modulo operation was implemented in MSVC and I tried to find something out. I don't really know assembly and all I had available was Express edition, without the source of VC\crt\src\intel\llrem.asm, but I think I managed to get some idea what's going on, after a bit of playing with the debugger and disassembly output. I tried to figure out how the remainder is calculated in case of positive integers and the divisor >=2^32. There is some code that deals with negative numbers of course, but I didn't dig into that.
Here is how I see it:
If divisor >= 2^32 both the dividend and the divisor are shifted right as much as necessary to fit the divisor into 32 bits. In other words: if it takes n digits to write the divisor down in binary and n > 32, n-32 least significant digits of both the divisor and the dividend are discarded. After that, the division is performed using hardware support for dividing 64 bit integers by 32 bit ones. The result might be incorrect, but I think it can be proved, that the result may be off by at most 1. After the division, the divisor (original one) is multiplied by the result and the product subtracted from the dividend. Then it is corrected by adding or subtracting the divisor if necessary (if the result of the division was off by one).
It's easy to divide 128 bit integer by 32 bit one leveraging hardware support for 64-bit by 32-bit division. In case the divisor < 2^32, one can calculate the remainder performing just 4 divisions as follows:
Let's assume the dividend is stored in:
DWORD dividend[4] = ...
the remainder will go into:
DWORD remainder;
1) Divide dividend[3] by divisor. Store the remainder in remainder.
2) Divide QWORD (remainder:dividend[2]) by divisor. Store the remainder in remainder.
3) Divide QWORD (remainder:dividend[1]) by divisor. Store the remainder in remainder.
4) Divide QWORD (remainder:dividend[0]) by divisor. Store the remainder in remainder.
After those 4 steps the variable remainder will hold what You are looking for.
(Please don't kill me if I got the endianess wrong. I'm not even a programmer)
In case the divisor is grater than 2^32-1 I don't have good news. I don't have a complete proof that the result after the shift is off by no more than 1, in the procedure I described earlier, which I believe MSVC is using. I think however that it has something to do with the fact, that the part that is discarded is at least 2^31 times less than the divisor, the dividend is less than 2^64 and the divisor is greater than 2^32-1, so the result is less than 2^32.
If the dividend has 128 bits the trick with discarding bits won't work. So in general case the best solution is probably the one proposed by GJ or caf. (Well, it would be probably the best even if discarding bits worked. Division, multiplication subtraction and correction on 128 bit integer might be slower.)
I was also thinking about using the floating point hardware. x87 floating point unit uses 80 bit precision format with fraction 64 bits long. I think one can get the exact result of 64 bit by 64 bit division. (Not the remainder directly, but also the remainder using multiplication and subtraction like in the "MSVC procedure"). IF the dividend >=2^64 and < 2^128 storing it in the floating point format seems similar to discarding least significant bits in "MSVC procedure". Maybe someone can prove the error in that case is bound and find it useful. I have no idea if it has a chance to be faster than GJ's solution, but maybe it's worth it to try.

Answer 8

The solution depends on what exactly you are trying to solve.
E.g. if you are doing arithmetic in a ring modulo a 64-bit integer then using
Montgomerys reduction is very efficient. Of course this assumes that you the same modulus many times and that it pays off to convert the elements of the ring into a special representation.
To give just a very rough estimate on the speed of this Montgomerys reduction: I have an old benchmark that performs a modular exponentiation with 64-bit modulus and exponent in 1600 ns on a 2.4Ghz Core 2. This exponentiation does about 96 modular multiplications (and modular reductions) and hence needs about 40 cycles per modular multiplication.

Answer 9

The accepted answer by #caf was real nice and highly rated, yet it contain a bug not seen for years.
To help test that and other solutions, I am posting a test harness and making it community wiki.
unsigned cafMod(unsigned A, unsigned B) {
assert(B);
unsigned X = B;
// while (X < A / 2) { Original code used <
while (X <= A / 2) {
X <<= 1;
}
while (A >= B) {
if (A >= X) A -= X;
X >>= 1;
}
return A;
}
void cafMod_test(unsigned num, unsigned den) {
if (den == 0) return;
unsigned y0 = num % den;
unsigned y1 = mod(num, den);
if (y0 != y1) {
printf("FAIL num:%x den:%x %x %x\n", num, den, y0, y1);
fflush(stdout);
exit(-1);
}
}
unsigned rand_unsigned() {
unsigned x = (unsigned) rand();
return x * 2 ^ (unsigned) rand();
}
void cafMod_tests(void) {
const unsigned i[] = { 0, 1, 2, 3, 0x7FFFFFFF, 0x80000000,
UINT_MAX - 3, UINT_MAX - 2, UINT_MAX - 1, UINT_MAX };
for (unsigned den = 0; den < sizeof i / sizeof i[0]; den++) {
if (i[den] == 0) continue;
for (unsigned num = 0; num < sizeof i / sizeof i[0]; num++) {
cafMod_test(i[num], i[den]);
}
}
cafMod_test(0x8711dd11, 0x4388ee88);
cafMod_test(0xf64835a1, 0xf64835a);
time_t t;
time(&t);
srand((unsigned) t);
printf("%u\n", (unsigned) t);fflush(stdout);
for (long long n = 10000LL * 1000LL * 1000LL; n > 0; n--) {
cafMod_test(rand_unsigned(), rand_unsigned());
}
puts("Done");
}
int main(void) {
cafMod_tests();
return 0;
}

Answer 10

As a general rule, division is slow and multiplication is faster, and bit shifting is faster yet. From what I have seen of the answers so far, most of the answers have been using a brute force approach using bit-shifts. There exists another way. Whether it is faster remains to be seen (AKA profile it).
Instead of dividing, multiply by the reciprocal. Thus, to discover A % B, first calculate the reciprocal of B ... 1/B. This can be done with a few loops using the Newton-Raphson method of convergence. To do this well will depend upon a good set of initial values in a table.
For more details on the Newton-Raphson method of converging on the reciprocal, please refer to http://en.wikipedia.org/wiki/Division_(digital)
Once you have the reciprocal, the quotient Q = A * 1/B.
The remainder R = A - Q*B.
To determine if this would be faster than the brute force (as there will be many more multiplies since we will be using 32-bit registers to simulate 64-bit and 128-bit numbers, profile it.
If B is constant in your code, you can pre-calculate the reciprocal and simply calculate using the last two formulae. This, I am sure will be faster than bit-shifting.
Hope this helps.

Answer 11

If 128-bit unsigned by 63-bit unsigned is good enough, then it can be done in a loop doing at most 63 cycles.
Consider this a proposed solution MSNs' overflow problem by limiting it to 1-bit. We do so by splitting the problem in 2, modular multiplication and adding the results at the end.
In the following example upper corresponds to the most significant 64-bits, lower to the least significant 64-bits and div is the divisor.
unsigned 128_mod(uint64_t upper, uint64_t lower, uint64_t div) {
uint64_t result = 0;
uint64_t a = (~0%div)+1;
upper %= div; // the resulting bit-length determines number of cycles required
// first we work out modular multiplication of (2^64*upper)%div
while (upper != 0){
if(upper&1 == 1){
result += a;
if(result >= div){result -= div;}
}
a <<= 1;
if(a >= div){a -= div;}
upper >>= 1;
}
// add up the 2 results and return the modulus
if(lower>div){lower -= div;}
return (lower+result)%div;
}
The only problem is that, if the divisor is 64-bits then we get overflows of 1-bit (loss of information) giving a faulty result.
It bugs me that I haven't figured out a neat way to handle the overflows.

Answer 12

I don't know how to compile the assembler codes, any help is appreciated to compile and test them.
I solved this problem by comparing against gmplib "mpz_mod()" and summing 1 million loop results. It was a long ride to go from slowdown (seedup 0.12) to speedup 1.54 -- that is the reason I think the C codes in this thread will be slow.
Details inclusive test harness in this thread:
https://www.raspberrypi.org/forums/viewtopic.php?f=33&t=311893&p=1873122#p1873122
This is "mod_256()" with speedup over using gmplib "mpz_mod()", use of __builtin_clzll() for longer shifts was essential:
typedef __uint128_t uint256_t[2];
#define min(x, y) ((x<y) ? (x) : (y))
int clz(__uint128_t u)
{
// unsigned long long h = ((unsigned long long *)&u)[1];
unsigned long long h = u >> 64;
return (h!=0) ? __builtin_clzll(h) : 64 + __builtin_clzll(u);
}
__uint128_t mod_256(uint256_t x, __uint128_t n)
{
if (x[1] == 0) return x[0] % n;
else
{
__uint128_t r = x[1] % n;
int F = clz(n);
int R = clz(r);
for(int i=0; i<128; ++i)
{
if (R>F+1)
{
int h = min(R-(F+1), 128-i);
r <<= h; R-=h; i+=(h-1); continue;
}
r <<= 1; if (r >= n) { r -= n; R=clz(r); }
}
r += (x[0] % n); if (r >= n) r -= n;
return r;
}
}

Answer 13

If you have a recent x86 machine, there are 128-bit registers for SSE2+. I've never tried to write assembly for anything other than basic x86, but I suspect there are some guides out there.

Answer 14

I am 9 years after the battle but here is an interesting O(1) edge case for powers of 2 that is worth mentioning.
#include <stdio.h>
// example with 32 bits and 8 bits.
int main() {
int i = 930;
unsigned char b = (unsigned char) i;
printf("%d", (int) b); // 162, same as 930 % 256
}

Answer 15

Since there is no predefined 128-bit integer type in C, bits of A have to be represented in an array. Although B (64-bit integer) can be stored in an unsigned long long int variable, it is needed to put bits of B into another array in order to work on A and B efficiently.
After that, B is incremented as Bx2, Bx3, Bx4, ... until it is the greatest B less than A. And then (A-B) can be calculated, using some subtraction knowledge for base 2.
Is this the kind of solution that you are looking for?