BCD List to HEX List

[John Machin]

Philippe Martin wrote:
> John Machin wrote:
>
> > Philippe Martin wrote:
> >> Hi,
> >>
> >> I'm looking for an algo that would convert a list such as:
> >
> > Such as what?
> >
> >>
> >> I'm using python to prototype the algo: this will move to C in an
> >> embedded system where an int has 16 bits - I do not wish to use any
> >> python library.
> >>
> >> l1 = [1,2,3,4,6,7,8] #represents the decimal number 12345678
> >
> > Does it??? How do you represent the decimal number 12349678, then?
> >
> >> l2 = func (l1)
> >> # l2 = [0x1, 0x2, 0xD, 0x6, 0x8, 0x7] #represents 0x12D687
> >>
> >
> > I'm sorry, but very little of that makes any sense to me:
> >
> > 1. I thought BCD meant something very much like this:
> > http://en.wikipedia.org/wiki/Binary-coded_decimal
> >
> > 2. >>> [0x1, 0x2, 0xD, 0x6, 0x8, 0x7] #represents 0x12D687
> > [1, 2, 13, 6, 8, 7]
> >
> > So [1], [2], [6] are unchanged, [3, 4] -> [13] (or maybe [3, 4, 5] ->
> > 13),  and [7, 8] -> [8,7].
> >
> > I doubt very much that there's an algorithm to do that. What is the
> > relationship between 1234(maybe 5)678 and 0x12D687??? I would expect
> > something like this::
> >
> >     0x12345678 (stored in 4 bytes 0x12, ..., 0x78)  -- or 0x21436587
> > or
> >     0x012345678s (where s is a "sign" nibble; stored in 5 bytes 0x01,
> > ..., 0x8s)
> >
> > IOW something regular and explicable ...
> >
> > 3. Perhaps it might be a good idea if you told us what the *real*
> > problem is, including *exact* quotes from the manual for the embedded
> > system. You evidently need/want to convert from one representation of
> > signed? unsigned? integers to another. Once we all understand *what*
> > those representations are, *then* we can undoubtedly help you with
> > pseudocode in the form of Python code manipulating lists or whatever.
> >
> > Cheers,
> > John
>
>
> Hi,
>
> From my answer to Marc:
>
> >My apologies, I clearly made a mistake with my calculator, yes the
> >resulting
> >array I would need is [0xb,0xc,0x6,0x1,0x4,0xe]
>

"Clearly"? I don't think that word means what you think it means :-)

All you need is something like the following. You will need to use
"long" if the C "int" is only 16 bits.

C:\junk>type bcd.py
def reconstitute_int(alist):
    reg = 0 # reg needs to be 32-bits (or more)
    for digit in alist:
        assert 0 <= digit <= 9
        reg = reg * 10 + digit
    return reg

def make_hex(anint):
    # anint needs to be 32-bits (or more)
    result = []
    while anint:
        result.append(anint & 0xF)
        anint >>= 4
    return result

def reverse_list(alist):
    n = len(alist)
    for i in xrange(n >> 1):
        reg1 = alist[n - 1 - i]
        reg2 = alist[i]
        alist[i] = reg1
        alist[n - 1 - i] = reg2

C:\junk>
C:\junk>python
Python 2.4.3 (#69, Mar 29 2006, 17:35:34) [MSC v.1310 32 bit (Intel)]
on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> import bcd
>>> num = bcd.reconstitute_int([1,2,3,4,5,6,7,8])
>>> num
12345678
>>> result = bcd.make_hex(num)
>>> result
[14, 4, 1, 6, 12, 11]
>>> bcd.reverse_list(result)
>>> result
[11, 12, 6, 1, 4, 14]
>>> ['0x%x' % digit for digit in result]
['0xb', '0xc', '0x6', '0x1', '0x4', '0xe']
>>> ^Z

HTH,
John