1. RE: Binary conversion
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
2. RE: Binary conversion
- Posted by rswiston at hotmail.com
Aug 15, 2002
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
>
>
>
>
3. RE: Binary conversion
I really wrote the last post in a moment of frustration.
What I didn't quite have a grasp on is the fact that if you do a poke4
of an atom with a value of 90 to memory and then do a peek({mem,4}) at
the same location you get {90,0,0,0}. The PC expert would say, "Well,
duh, a PC stores binary numbers with significant digit first!" (or
last...no first...oh, whatever). What I've been working on is wrappers
for DLLs that interface to an IBM iSeries. IBM's method is to store
significant digits last so your 90 would look like {0,0,0,90}. I tried
doing a value() against my {0,0,0,90} which returned nothing (as it
should).
So that's why I wondered about binary conversion. Still wondering on
how to turn {0,0,0,90} back into 90. I'm gonna keep diggin'.
Jonas
4. RE: Binary conversion
>So that's why I wondered about binary conversion. Still wondering on
>how to turn {0,0,0,90} back into 90. I'm gonna keep diggin'.
Did you try this:
include machine.e
include misc.e
constant binary = {0, 0, 0, 90}
sequence rev
integer answer
-- reverse bytes so they will be 386 order
rev = reverse(binary)
-- convert to an integer
answer = bytes_to_int(rev)
5. RE: Binary conversion
What a great community this is! I really appreciate everyone's
willingness to help.
Here's a little more background to my problem and the solution I came up
with.
I have an atom with a value of 90. If you were to poke4() this atom to
memory and then peek4() you would get {90,0,0,0}. Since the iSeries
expects the reverse order I called a translation routine to reverse the
order before passing to the iSeries. Everything worked up to that
point. When the data was returned to my PC is was {0,0,0,91}
(remembering that I reversed it and the program on the iSeries added 1
to the number).
I liked the reverse() and bytes_to_int() solution and it worked, but not
on negative values (the doc states that bytes_to_int expects the
sequence to contain positive values). I thought about Derek's
multiplication approach but here's what I came up with:
atom bin, parm
sequence rtn_seq
...some stuff in here...
bin = allocate(4)
rtn_seq = reverse(rtn_seq)
poke(bin, rtn_seq)
parm = peek4s(bin)
free(bin)
So what I accomplished was I used reverse() to put the sequence back
into the order needed for an atom, poked the sequence to memory and then
used peek4s to get the value back out. Maybe not the most elegant
solution but it worked nonetheless.
Thanks again to everyone!
6. RE: Binary conversion
> 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
Hey Jonas,
I saw the other posts, but I am not real sure about what exactly
it is you are trying to do.
poke4 is a 32 bit command so you will store 4 bytes of information
at a specified location. If this is the case, I do not see why
you are using peek to get the information back out. Peek is an 8
bit command so, as posted, you would get back a 4 length sequence
representing the 4 bytes that you wrote out earlier.
Why not use peek4u or peek4s to read the data back out as these
are the 32 bit routines for accessing memory?
Write 90: poke4( mem, 90 )
Read 90: peek4u( mem )
Don
7. RE: Binary conversion
Don,
The problem was that the IBM iSeries stores 32 bit values in the
opposite order of the PC. What I wound up with after a call to a
program on the iSeries is 4 bytes returned to me in the opposite order.
You can't use peek4s or peek4u on the memory since peek4s/u is expecting
the bytes to be in the PC order.
Make sense?
Don Phillips wrote:
> Hey Jonas,
> I saw the other posts, but I am not real sure about what exactly
> it is you are trying to do.
>
> poke4 is a 32 bit command so you will store 4 bytes of information
> at a specified location. If this is the case, I do not see why
> you are using peek to get the information back out. Peek is an 8
> bit command so, as posted, you would get back a 4 length sequence
> representing the 4 bytes that you wrote out earlier.
>
> Why not use peek4u or peek4s to read the data back out as these
> are the 32 bit routines for accessing memory?
>
> Write 90: poke4( mem, 90 )
> Read 90: peek4u( mem )
>
> Don
>
>
8. RE: Binary conversion
> Don,
>
> The problem was that the IBM iSeries stores 32 bit values in the
> opposite order of the PC. What I wound up with after a call to a
> program on the iSeries is 4 bytes returned to me in the opposite order.
>
> You can't use peek4s or peek4u on the memory since peek4s/u is expecting
>
> the bytes to be in the PC order.
>
> Make sense?
Ya, complete. Thats just a pain in the rump =), looks like I will bow
out of this thread. Source already posted on the topic looks efficient
enough that I dont want to play with it heh.
Don
