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 -- http://mail.python.org/mailman/listinfo/python-list