在 C/C++ 中使用 % (模数)有什么替代方法吗? [英] Is there any alternative to using % (modulus) in C/C++?

问题描述

I read somewhere once that the modulus operator is inefficient on small embedded devices like 8 bit micro-controllers that do not have integer division instruction. Perhaps someone can confirm this but I thought the difference is 5-10 time slower than with an integer division operation.

Is there another way to do this other than keeping a counter variable and manually overflowing to 0 at the mod point?

const int FIZZ = 6;
for(int x = 0; x < MAXCOUNT; x++)
{
    if(!(x % FIZZ)) print("Fizz
"); // slow on some systems
}

vs:

The way I am currently doing it:

const int FIZZ = 6;
int fizzcount = 1;
for(int x = 1; x < MAXCOUNT; x++)
{
    if(fizzcount >= FIZZ) 
    {
        print("Fizz
");
        fizzcount = 0;
    }
}

解决方案

Ah, the joys of bitwise arithmetic. A side effect of many division routines is the modulus - so in few cases should division actually be faster than modulus. I'm interested to see the source you got this information from. Processors with multipliers have interesting division routines using the multiplier, but you can get from division result to modulus with just another two steps (multiply and subtract) so it's still comparable. If the processor has a built in division routine you'll likely see it also provides the remainder.

Still, there is a small branch of number theory devoted to Modular Arithmetic which requires study if you really want to understand how to optimize a modulus operation. Modular arithmatic, for instance, is very handy for generating magic squares.

So, in that vein, here's a very low level look at the math of modulus for an example of x, which should show you how simple it can be compared to division:

---

Maybe a better way to think about the problem is in terms of number bases and modulo arithmetic. For example, your goal is to compute DOW mod 7 where DOW is the 16-bit representation of the day of the week. You can write this as:

 DOW = DOW_HI*256 + DOW_LO

 DOW%7 = (DOW_HI*256 + DOW_LO) % 7
       = ((DOW_HI*256)%7  + (DOW_LO % 7)) %7
       = ((DOW_HI%7 * 256%7)  + (DOW_LO%7)) %7
       = ((DOW_HI%7 * 4)  + (DOW_LO%7)) %7

Expressed in this manner, you can separately compute the modulo 7 result for the high and low bytes. Multiply the result for the high by 4 and add it to the low and then finally compute result modulo 7.

Computing the mod 7 result of an 8-bit number can be performed in a similar fashion. You can write an 8-bit number in octal like so:

  X = a*64 + b*8 + c

Where a, b, and c are 3-bit numbers.

  X%7 = ((a%7)*(64%7) + (b%7)*(8%7) + c%7) % 7
      = (a%7 + b%7 + c%7) % 7
      = (a + b + c) % 7

since 64%7 = 8%7 = 1

Of course, a, b, and c are

  c = X & 7
  b = (X>>3) & 7
  a = (X>>6) & 7  // (actually, a is only 2-bits).

The largest possible value for a+b+c is 7+7+3 = 17. So, you'll need one more octal step. The complete (untested) C version could be written like:

unsigned char Mod7Byte(unsigned char X)
{
    X = (X&7) + ((X>>3)&7) + (X>>6);
    X = (X&7) + (X>>3);

    return X==7 ? 0 : X;
}

I spent a few moments writing a PIC version. The actual implementation is slightly different than described above

Mod7Byte:
       movwf        temp1        ;
       andlw        7        ;W=c
       movwf        temp2        ;temp2=c
       rlncf   temp1,F        ;
       swapf        temp1,W ;W= a*8+b
       andlw   0x1F
       addwf        temp2,W ;W= a*8+b+c
       movwf        temp2   ;temp2 is now a 6-bit number
       andlw   0x38    ;get the high 3 bits == a'
       xorwf        temp2,F ;temp2 now has the 3 low bits == b'
       rlncf   WREG,F  ;shift the high bits right 4
       swapf   WREG,F  ;
       addwf        temp2,W ;W = a' + b'

 ; at this point, W is between 0 and 10


       addlw        -7
       bc      Mod7Byte_L2
Mod7Byte_L1:
       addlw        7
Mod7Byte_L2:
       return

Here's a liitle routine to test the algorithm

       clrf    x
       clrf    count

TestLoop:
       movf        x,W
       RCALL   Mod7Byte
       cpfseq count
        bra    fail

       incf        count,W
       xorlw   7
       skpz
        xorlw        7
       movwf   count

       incfsz        x,F
       bra        TestLoop
passed:

Finally, for the 16-bit result (which I have not tested), you could write:

uint16 Mod7Word(uint16 X)
{
 return Mod7Byte(Mod7Byte(X & 0xff) + Mod7Byte(X>>8)*4);
}

Scott

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