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How to get the negate of decimal to binary?

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# 1  
Old 12-11-2013
How to get the negate of decimal to binary?

Hi All,
New to this forum (and yes , a newbie in programming..Smilie)

I have a decimal to binary converter script done this way :


bit0=$(( (i & 0x01) > 0 ))
bit1=$(( (i & 0x02) > 0 ))
bit2=$(( (i & 0x04) > 0 ))
bit3=$(( (i & 0x08) > 0 ))
bit4=$(( (i & 0x10) > 0 ))
bit5=$(( (i & 0x20) > 0 ))
bit6=$(( (i & 0x40) > 0 ))
bit7=$(( (i & 0x80) > 0 ))

Fisrt I would like to understand how the expression $(( (i & 0x01) > 0 )) works

Next, I need to obtain the same bits to its negate, so i.e. if for decimal 177


I need


is there a way to accomplish this inside the same script ??

Thank you very much.
# 2  
Old 12-11-2013
Are you sure it is a negate or an INVERT?

What you are showing looks like an INVERT...

So just subtract 177 from 255 and voila you have your decimal 78...

BTW is this homework?

Last edited by wisecracker; 12-11-2013 at 05:25 AM.. Reason: Corect my error...
# 3  
Old 12-11-2013

I never seen the thing under this aspect...

Effectively, let's take 98 :


do 255-98=157 that is


comparing the two:


Exactly the opposite cosidering bit to bit !!!!! Smilie

So I should keep the first part of the script to obtain the first conversion and add a second part to do 255-inputvalue and then again another conversion



bit0=$(( (i & 0x01) > 0 ))
bit1=$(( (i & 0x02) > 0 ))
bit2=$(( (i & 0x04) > 0 ))
bit3=$(( (i & 0x08) > 0 ))
bit4=$(( (i & 0x10) > 0 ))
bit5=$(( (i & 0x20) > 0 ))
bit6=$(( (i & 0x40) > 0 ))
bit7=$(( (i & 0x80) > 0 ))

bit8=$(( (n & 0x01) > 0 ))
bit9=$(( (n & 0x02) > 0 ))
bitA=$(( (n & 0x04) > 0 ))
bitB=$(( (n & 0x08) > 0 ))
bitC=$(( (n & 0x10) > 0 ))
bitD=$(( (n & 0x20) > 0 ))
bitE=$(( (n & 0x40) > 0 ))
bitF=$(( (n & 0x80) > 0 ))


Is there perhaps a quicker negation of already producted single bit0-bit7 bits ??

I need it for a little project in BeagleBone GPIO, not quite homework...(homework should be done many years ago....when I was younger with less woody brain...Smilie )

Thank you

Last edited by digiteltlc; 12-11-2013 at 05:44 AM..
# 4  
Old 12-11-2013

REMEBER! What I have shown you is for 8 bits...
IF you require 16 bits the subtract your number from 65535...
(Similarly for 32 bits, etc...)

Look at your problem logically...

"$1" is the first user argument of your shell script and it looks as though you will be entering your script as something like this:-
Your prompt> somescript 10101010<CR>

<CR> == carraiage return...

You have saved this argument inside your variable "i"

So "i" now contains a _string_ representation of your binary value...

Use something like:-

To change each bit...

This method keeps it all inside bash builtins...

I am at work ATM so replies will be sporadic from me so have a go yourself... ;o)
# 5  
Old 12-11-2013
The argument passed to script is a decimal number
Anyway the method you suggested me (255-i then reconversion) works 100%
Just tested on real circuit.

So i'll go that way

Thank you very much again
# 6  
Old 12-11-2013
Might I suggest using what is already available?
$ i=98
$ typeset -Z8 n=`echo "ibase=10;obase=2;255-$i"|bc`
$ echo $n

I hope that this is useful. I got the basis of this elsewhere on the board so I thought I should share it.

# 7  
Old 12-11-2013
!!! very good as well.. !!!!

Thank you very much

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