Tony,

I hate to do this to you, but here is simpler(?) way to round to nearest 5:

Given 25 bit number HI:MID:LO

Its value is

LO + 256*MID + 65536*HI

Which is equal to

LO+255*MID+MID+65535*HI+HI

But 255*MID and 65535*HI are both multiples of 5, so:

LO+MID+HI == HI:MID:LO (mod 5)

This can be at most a 10 bit number, let us call it H2:L2

now we can:

    movf    L2,W,A
    andlw  0x0F
    movwf ONES,A

    swapf L2,W,A
    andlw 0x0F
    btfsc   L2,4
    addlw 5        ; worth 6, not 1

    btfsc    H2,0
    addlw    6
    btfsc    H2,1
    addlw    2

and Bob's your uncle!

Bob Ammerman
RAm Systems
(contract development of high performance, high function, low-level
software)

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