RE: Binary conversion
- Posted by rswiston at hotmail.com Aug 15, 2002
- 427 views
One way might be to do this: atom binary, sum binary = {0,0,0,2} sum = 0 for x = 1 to length(binary) do sum += (power(10, length(binary)-x))*binary[x] end for this would convert any langth of binary into a decimal number (replace ten within the loop to convert to a different base) so 0,0,0,2 = 0*10^3 + 0*10^2 + 0*10^1 + 2*10^0 = 2 additionally.. 1,2,3,4,5,6 = 1*10^5 + 2*10^4 + 3*10^3 + 4*10^2 + 5*10^1 + 6*10^0 = 12345 Note: This will only change a sequence of "digits" into one atom (or number), not convert from one base into another. >From: rudy toews <rltoews at ilos.net> >Reply-To: EUforum at topica.com >To: EUforum <EUforum at topica.com> >Subject: RE: Binary conversion >Date: Thu, 15 Aug 2002 22:41:17 +0000 > > >Jonas Temple wrote: > > Okay, I asked this a while back and now I've got a slightly different > > problem. > > > > If I have a sequence of {0,0,0,2} that is supposed to represent a binary > > > > number, how do I convert that to an atom? > > > > Jonas > > > > >hi Jonas >i did not see the orginal posting. >but would a table be helpful? >since the smallest element in Euphoria is an atom then a table of 32 >entrys of the 32 different bit positions would be needed. > >using 'or', 'and' or 'xor' should work > >receiver = receiver or'd with table entry(input position) > >this is my first thought on it. working nights makes one groggy. >later >rudy > > >lotterywars > > > >