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\n"); // 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\n");
fizzcount = 0;
}
}
```

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

If you are calculating a number mod some power of two, you can use the bit-wise and operator. Just subtract one from the second number. For example:

```
x % 8 == x & 7
x % 256 == x & 255
```

A few caveats:

- This
**only works**if the second number is a power of two. - It's only equivalent if the modulus is always positive. The C and C++ standards don't specify the sign of the modulus when the first number is negative (until C++11, which
*does*guarantee it will be negative, which is what most compilers were already doing). A bit-wise and gets rid of the sign bit, so it will always be positive (i.e. it's a true modulus, not a remainder). It sounds like that's what you want anyway though. **Your compiler probably already does this when it can, so in most cases it's not worth doing it manually.**

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