Fast modulo 10 in c | 易学教程

问题

I am looking for a fast modulo 10 algorithm because I need to speed up my program which does many modulo operations in cycles.

I have checked out this page which compares some alternatives. As far as I understand it correctly, T3 was the fastest of all. My question is, how would x % y look like using T3 technique?

I copied T3 technique here for simplicity in case the link gets down.

for (int x = 0; x < max; x++)
{
        if (y > (threshold - 1))
        {
               y = 0; //reset
               total += x;
        }
        y += 1;
}

Regarding to comments, if this is not really faster then regular mod, I am looking for at least 2 times faster modulo than using %. I have seen many examples with use power of two, but since 10 is not, how can I get it to work?

Edit:

For my program, let's say I have 2 for cycles where n=1 000 000 and m=1000.

Looks like this:

for (i = 1; i <= n; i++) {
        D[(i%10)*m] = i;
        for (j = 1; j <= m; j++) {
           ...
        }
}

回答1:

Here's the fastest modulo-10 function you can write:

unsigned mod10(unsigned x)
{
    return x % 10;
}

And here's what it looks like once compiled:

movsxd rax, edi
imul rcx, rax, 1717986919
mov rdx, rcx
shr rdx, 63
sar rcx, 34
add ecx, edx
add ecx, ecx
lea ecx, [rcx + 4*rcx]
sub eax, ecx
ret

Note the lack of division/modulus instructions, the mysterious constants, the use of an instruction which was originally intended for complex array indexing, etc. Needless to say, the compiler knows a lot of tricks to make your program as fast as possible. You'll rarely beat it on tasks like this.

回答2:

You likely can't beat the compiler.

Debug build

//     int foo = x % 10;
010341C5  mov         eax,dword ptr [x]  
010341C8  cdq  
010341C9  mov         ecx,0Ah  
010341CE  idiv        eax,ecx  
010341D0  mov         dword ptr [foo],edx

Retail build (doing some ninja math there...)

//    int foo = x % 10;
00BD100E  mov         eax,66666667h  
00BD1013  imul        esi  
00BD1015  sar         edx,2  
00BD1018  mov         ecx,edx  
00BD101A  shr         ecx,1Fh  
00BD101D  add         ecx,edx  
00BD101F  lea         eax,[ecx+ecx*4]  
00BD1022  add         eax,eax  
00BD1024  sub         esi,eax

回答3:

The code isn’t a direct substitute for modulo, it substitutes modulo in that situation. You can write your own mod by analogy (for a, b > 0):

int mod(int a, int b) {
    while (a >= b) a -= b;
    return a;
}

… but whether that’s faster than % is highly questionable.

回答4:

This will work for (multiword) values larger than the machineword (but assuming a binary computer ...):

#include <stdio.h>

unsigned long mod10(unsigned long val)
{
unsigned res=0;

res =val &0xf;
while (res>=10) { res -= 10; }

for(val >>= 4; val; val >>= 4){
        res += 6 * (val&0xf);
        while (res >= 10) { res -= 10; }
        }

return res;
}

int main (int argc, char **argv)
{
unsigned long val;
unsigned res;

sscanf(argv[1], "%lu", &val);

res = mod10(val);
printf("%lu -->%u\n", val,res);

return 0;
}

UPDATE: With some extra effort, you could get the algoritm free of multiplications, and with the proper amount of optimisation we can even get the recursive call inlined:

static unsigned long mod10_1(unsigned long val)
{
unsigned char res=0; //just to show that we don't need a big accumulator

res =val &0xf; // res can never be > 15
if (res>=10) { res -= 10; }

for(val >>= 4; val; val >>= 4){
        res += (val&0xf)<<2 | (val&0xf) <<1;
        res= mod10_1(res); // the recursive call
        }

return res;
}

And the result for mod10_1 appears to be mul/div free and almost without branches:

mod10_1:
.LFB25:
    .cfi_startproc
    movl    %edi, %eax
    andl    $15, %eax
    leal    -10(%rax), %edx
    cmpb    $10, %al
    cmovnb  %edx, %eax
    movq    %rdi, %rdx
    shrq    $4, %rdx
    testq   %rdx, %rdx
    je      .L12
    pushq   %r12
    .cfi_def_cfa_offset 16
    .cfi_offset 12, -16
    pushq   %rbp
    .cfi_def_cfa_offset 24
    .cfi_offset 6, -24
    pushq   %rbx
    .cfi_def_cfa_offset 32
    .cfi_offset 3, -32
.L4:
    movl    %edx, %ecx
    andl    $15, %ecx
    leal    (%rcx,%rcx,2), %ecx
    leal    (%rax,%rcx,2), %eax
    movl    %eax, %ecx
    movzbl  %al, %esi
    andl    $15, %ecx
    leal    -10(%rcx), %r9d
    cmpb    $9, %cl
    cmovbe  %ecx, %r9d
    shrq    $4, %rsi
    leal    (%rsi,%rsi,2), %ecx
    leal    (%r9,%rcx,2), %ecx
    movl    %ecx, %edi
    movzbl  %cl, %ecx
    andl    $15, %edi
    testq   %rsi, %rsi
    setne   %r10b
    cmpb    $9, %dil
    leal    -10(%rdi), %eax
    seta    %sil
    testb   %r10b, %sil
    cmove   %edi, %eax
    shrq    $4, %rcx
    andl    $1, %r10d
    leal    (%rcx,%rcx,2), %r8d
    movl    %r10d, %r11d
    leal    (%rax,%r8,2), %r8d
    movl    %r8d, %edi
    andl    $15, %edi
    testq   %rcx, %rcx
    setne   %sil
    leal    -10(%rdi), %ecx
    andl    %esi, %r11d
    cmpb    $9, %dil
    seta    %bl
    testb   %r11b, %bl
    cmovne  %ecx, %edi
    andl    $1, %r11d
    andl    $240, %r8d
    leal    6(%rdi), %ebx
    setne   %cl
    movl    %r11d, %r8d
    andl    %ecx, %r8d
    leal    -4(%rdi), %ebp
    cmpb    $9, %bl
    seta    %r12b
    testb   %r8b, %r12b
    cmovne  %ebp, %ebx
    andl    $1, %r8d
    cmovne  %ebx, %edi
    xorl    $1, %ecx
    andl    %r11d, %ecx
    orb     %r8b, %cl
    cmovne  %edi, %eax
    xorl    $1, %esi
    andl    %r10d, %esi
    orb     %sil, %cl
    cmove   %r9d, %eax
    shrq    $4, %rdx
    testq   %rdx, %rdx
    jne     .L4
    popq    %rbx
    .cfi_restore 3
    .cfi_def_cfa_offset 24
    popq    %rbp
    .cfi_restore 6
    .cfi_def_cfa_offset 16
    movzbl  %al, %eax
    popq    %r12
    .cfi_restore 12
    .cfi_def_cfa_offset 8
    ret
.L12:
    movzbl  %al, %eax
    ret
    .cfi_endproc
.LFE25:
    .size   mod10_1, .-mod10_1
    .p2align 4,,15
    .globl  mod10
    .type   mod10, @function

来源：https://stackoverflow.com/questions/50066237/fast-modulo-10-in-c

标签

algorithm

optimization

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