Shell Scripts and bitwise operator

This doesn't work on /bin/sh

Try

let 1 \| 2
awk '{print or(1, 2)}'
gawk '{print or(1, 2)}'

Why not use ksh?

the sh syntax also work on ksh, and ksh support:

echo echo $(( 5 & 6 ))
echo $(( 5 | 6 ))

all you need to do is to change the first line of your script
to:

#!/bin/ksh

or
#!/usr/bin/ksh

it depends on where your ksh is located.

Happy New Year to all of you !

There are a couple of options in bourne shell.  

#!/bin/sh

num=7

if [ "$num" -ge 6 ] && [ "$num" -le 8 ] ; then
        echo "The first AND statement works."
else
        echo "The first AND statement failed."
fi

if [ "$num" -ge 6 -a "$num" -le 8 ] ; then
        echo "The second AND statement also works."
else
        echo "The second AND statement also failed."
fi

if [ "$num" -eq 7 -o "$num" -eq 8 ] ; then
        echo "The first OR statement works."
else
        echo "The first OR statement failed."
fi

if [ "$num" -eq 7 ] || [ "$num" -eq 8 ] ; then
        echo "The second OR statement also works"
else
        echo "The second OR statement also failed."
fi

I am working with a old script and it uses sh.
So, I will not be able to change it to ksh or bash.

Also, I don't want to use perl. (I am working on something related to OS and nobody has used perl till now and hence I also don't want to use it)

So, none of the solutions suggested till now works for me.
Is there any other solution other than what is suggested till now?

#!/bin/sh
bash -c 'echo $(( 5|6 ))'

#!/bin/sh
csh -c '@ x = ( 5 & 6 ) ; echo $x'

Thanks for this oz.
I will remind again. I am not allowed to use bash, csh or perl inside this script.

Even in case, I use bash (or perl),  your suggestion is not working.

Here is the sample script that I have:
#!/bin/sh

nm1=ff
N1=10
F1=`bash -c 'echo $(( $N1&$nm1 ))'`
#F1=`perl -e 'print $N1&$nm1'`
echo F1 = $F1

This throws error.
Here, N1 is decimal and nm1 is hexadecimal.

You have any corrections for this?

Thanks,
Narendra

F1=`bash -c "echo $(( $N1 & 0x$nm1 ))"`

F1=`echo 'main(){ printf("%d",'"$N1&0x$nm1);}" > $$.c ; cc $$.c ; ./a.out`

What is it about my example that didn't work for you?  It's written solely in bourne shell (or, sh as you like to call it).  There aren't any references to ksh, bash, or perl.  Am I missing something?  I thought that answered your question.

Hi ozo,

    I liked all your solutions and all of them work.
    Hats off to all those ideas.

    I will keep this thread for some more time.
    So, that more new ideas can flow.

Thanks,
Narendra

Hello devrick0,

    I think, you didn't understand the question properly.
    I am not looking for Logical AND/OR operations.

    I am looking for 'BITWISE' AND/OR operations.

    Suppose I do AND of 5 and 6. The result has to be 4.
    If I do OR of 5 and 6. The result has to be 7.

    So, does your code do this?

    Hope you got my point!?

> It's written solely in bourne shell (or, sh as you like to call it).
    Bourne shell is not sh. It is bash. Both are different (though many times they look very similar)

Thanks,
Narendra

>  I am using /bin/sh ..
traditional sh has no bitwise operators

> ..  I am not allowed to use bash, csh or perl inside this script.
can you use some other tools?
  echo 0|gawk '{print and(5,6)}'
  echo 0|gawk '{print or(5,6)}'

>  Bourne shell is not sh. It is bash.
wrong
sh is Bourne shell, bash is born/Bourne again shell

I stand corrected (on the bitwise portion of my response).  I had overlooked that, saw the AND/OR comparisons in bourne shell, and went with it.

For future reference (excerpt from O'Reilly's UNIX Power Tools book).
sh:
The Bourne shell (named after its creator, Steve Bourne). This is the oldest of the current UNIX shells
and is available on most UNIX systems. (Some systems have replaced sh with a newer shell, like ksh
or bash, that has the features of sh and more.) It is a bit primitive and lacks job control features (the
ability to move jobs from the foreground to the background). Most UNIX users consider the Bourne
shell superior for shell programming or writing command files.

bash:
The "Bourne−again" shell developed by the Free Software Foundation (52.9). bash (8.2) is fairly
similar to the Korn shell. It has many of the C shell's features, plus history editing and a built−in help
command.

Hi ahoffmann,

> echo 0|gawk '{print and(5,6)}'
I don't have gawk. But, I can use awk or nawk.
And I am getting this error (with nawk):
nawk: calling undefined function and
 input record number 1
 source line number 1

Regards,
Narendra

Hi devrick0,

    Thanks for pointing to reference in O'Reilly.
    As, I said before, it is also telling that sh and bash are different. (That's the reason they have different executables)

    All the features of sh may work in bash and bash is very good, user friendly and also very powerfull.
    But, all the features of bash will not work in sh!

    Anyway, the solution suggested by ozo is working.
    The only problem is they are using either bash or perl or creating temporary files.

    I am waiting to see, if there are any other alternative solutions for this.
    Please do suggest.

Thanks,
Narendra

nawk also has no bitwise operators
I'd write my own simple AND() and OR() function in awk.

You could also write simple AND and OR programs in C, and then invoke those functions in your  shell script if you don't want to invoke bash, ksh, csh, perl, ruby, etc.

Bitwise operators implemented with only Bourne Shell builtins, expr, and cut.  Please tell me my usage of expr and cut are POSIX compliant...

#!/bin/sh

HIGHEST_BIT_MASK=2147483648           # 32-bit (10000000000000000000000000000000)

# [/test builtin barfs on this.  expr here on Linux seems to support 64 bit though.
#HIGHEST_BIT_MASK=9223372036854775808  # 64-bit (1000000000000000000000000000000000... you get the idea)

bit_pairs_op() {
    if [ $# -ne 3 ]; then
        echo "Usage:  bit_pairs_op <num1> <op> <num2>" >&2
        exit 0
    fi

    NUM1=$1
    NUM2=$3
    OP=$2
    MASK=$HIGHEST_BIT_MASK
    while [ $MASK -gt 0 ]; do
        BIT1=`expr $NUM1 / $MASK`
        if [ $BIT1 -eq 1 ]; then
            NUM1=`expr $NUM1 - $MASK`
        fi

        BIT2=`expr $NUM2 / $MASK`
        if [ $BIT2 -eq 1 ]; then
            NUM2=`expr $NUM2 - $MASK`
        fi

        expr $BIT1 $OP $BIT2

        MASK=`expr $MASK / 2`
    done
}

binary_to_decimal() {
    if [ $# -ne 1 ]; then
        echo "Usage:  binary_to_decimal <binary>" >&2
        exit 0
    fi

    BINARY=$1
    MASK=$HIGHEST_BIT_MASK
    BITNUM=1
    DECIMAL=0
    while [ $MASK -gt 0 ]; do
        BIT=`echo $BINARY | cut -c$BITNUM`
        if [ $BIT -eq 1 ]; then
            DECIMAL=`expr $DECIMAL + $MASK`
        fi

        MASK=`expr $MASK / 2`
        BITNUM=`expr $BITNUM + 1`
    done

    echo $DECIMAL
}

or() {
    if [ $# -ne 2 ]; then
        echo "Usage:  op <num1> <num2>" >&2
        exit 0
    fi

    eval `bit_pairs_op $1 + $2 | or_result`

    echo "$RESULT_BITS"
}

or_result() {
    RESULT_BITS=""
    while read RESULT; do
        if [ $RESULT -gt 0 ]; then
            RESULT_BITS="${RESULT_BITS}1"
        else
            RESULT_BITS="${RESULT_BITS}0"
        fi
        echo "RESULT_BITS=$RESULT_BITS"
    done
}

and() {
    if [ $# -ne 2 ]; then
        echo "Usage:  and <num1> <num2>" >&2
        exit 0
    fi

    eval `bit_pairs_op $1 + $2 | and_result`

    echo "$RESULT_BITS"
}

and_result() {
    RESULT_BITS=""
    while read RESULT; do
        if [ $RESULT -eq 2 ]; then
            RESULT_BITS="${RESULT_BITS}1"
        else
            RESULT_BITS="${RESULT_BITS}0"
        fi
        echo "RESULT_BITS=$RESULT_BITS"
    done
}

xor() {
    if [ $# -ne 2 ]; then
        echo "Usage:  and <num1> <num2>" >&2
        exit 0
    fi

    eval `bit_pairs_op $1 - $2 | xor_result`

    echo "$RESULT_BITS"
}

xor_result() {
    RESULT_BITS=""
    while read RESULT; do
        if [ $RESULT -eq -1 ]; then
            RESULT_BITS="${RESULT_BITS}1"
        else
            RESULT_BITS="${RESULT_BITS}$RESULT"
        fi
        echo "RESULT_BITS=$RESULT_BITS"
    done
}

if [ $# -ne 3 ]; then
    echo "Usage: $0 <num1> <op> <num2>"
    exit 1
fi

binary_to_decimal `$2 $1 $3`

The above was done for fun.  It is mighty slow ;)

Problem lies in conversion from HEX to decimal .
Use bc to get it done and use sh -c instead of bash/ksh/perl if you orig script is in "sh".

#!/bin/sh
#Since bc accepts A-F only not a-f
nm1=FF
N1=10
#That's it convert Hex into decimal
nm2=`expr "ibase=16; $nm1;" | bc`
#voila got it ....!
F1=`sh -c 'echo $(( $N1&$nm2 ))'`

Thanks
Dinesh

I'd use awk's printf to convert types, much faster than piping through sevaral tools :)

Nope
 no inbuilt conversion from hex to decimal is possible , as far as awk is concerned.
Try following
echo "FF" | awk '{ printf "%d",$0 }'
and
echo "55" | awk '{ printf "%x",$0 }'

First gives 0, second gives 37.
As soon as awk encounters any alphabet 'A-F','a-f',0x,
It is giving 0.
I have checked it.
Thanks
Dinesh

echo ''|awk '{print 0xff }'
echo ''|awk '{print 0xFf -3 }'

oops, last suggestion is gawk only, sorry.

I am accepting the answers gave by ozo.
Only they were near to the solution.

But, it also didn't work out.
I don't know if many people understand what it means to write a code in the operating system.
We will have lot of constraints.
We will not be having so many shells. Only sh is available.
So, using ksh or bash or csh is ruled out.
And using perl or awk or any other tool means adding a dependency. Nobody in OS programming will accept dependency.
The question that will be asked is, what if we don't have the tool? Then our solution will fail!!

That's the reason I didn't give A for the soltuion.
I am very sorry for that.

Regards,
Narendra