```;*******************************************************************
;scale_hex2dec
;  The purpose of this routine is to scale a hexadecimal byte to a
;decimal byte. In other words, if 'h' is a hexadecimal byte then
;the scaled decimal equivalent 'd' is:
;    d = h * 100/256.
;Note that this can be simplified:
;    d = h * 25 / 64 = h * 0x19 / 0x40
;Multiplication and division can be expressed in terms of shift lefts
;and shift rights:
;    d = [ (h<<4) + (h<<3) + h ] >> 6
;The program divides the shifting as follows so that carries are
automatically
;taken care of:
;    d =   (h + (h + (h>>3)) >> 1) >> 2
;
;Inputs:   W - should contain 'h', the hexadecimal value to be scaled
;Outputs:  W - The scaled hexadecimal value is returned in W
;Memory:   temp
;Calls:    none
scale_hex2dec
MOVWF   temp            ;Hex value is in W.
CLRC                    ;Clear the Carry bit so it doesn't affect
RRF
RRF     temp,F
CLRC
RRF     temp,F
CLRC
RRF     temp,F          ;temp = h>>3
ADDWF   temp,F          ;temp = h + (h>>3)
RRF     temp,F          ;temp = (h + (h>>3)) >> 1
ADDWF   temp,F          ;temp = h + ((h + (h>>3)) >> 1)
RRF     temp,F
CLRC
RRF     temp,W          ;d = W = (h + (h + (h>>3)) >> 1) >> 2
RETURN

```

For the more general case:

``` C = A*B/255
```

There is another trick that you can attempt. Recall the power series for division:

```   N        N    /     / e \   / e \2  / e \3      \
-------  = --- * | 1 - |---| + |---| - |---| + ... |
v + e      v    \     \ v /   \ v /   \ v /       /
```

If, N=A*B, v+e = 255. If you let v=256 and e=-1 then the series simplifies to:

```   A*B      A*B    /    / 1 \   / 1 \2  / 1 \3      \
--------  = --- * | 1 + |---| + |---| + |---| + ... |
256 - 1    256    \    \256/   \256/   \256/       /
```

Or just keeping the first two terms:

```   A*B       A*B    /     1  \
--------  ~= --- * | 1 + --- |
255       256    \    256 /
```

If A & B are 8-bit quantites, then the second term in the series would produce 0 (you'd be dividing the product A*B by 256 twice). However, as Harold notes, you may use the second term to round the result. Specifically, if you treated the multiplication (conceptually) as floating point then you'll end up with a fractional component that's greater than 0.5 if A*B is greater than 2^15. Or stated differently, if the most significant bit is set in the product, then increment the result.

 file: /Techref/microchip/scale.htm, 2KB, , updated: 1999/9/28 16:29, local time: 2018/11/18 22:15, TOP NEW HELP FIND:  54.90.185.120:LOG IN

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