Show Hexadecimal Numbers Of a File

Question

I want to build a bash program that can read a file, like a *.bin and print all its hexadecimal numbers, as 'hex' editors do. Where I can start?

Answer 1

Use the od command,

od -t x1  filename

Answer 2

You could use od. "od -x file" Why reinvent that wheel?

Answer 3

you can also use hexdump if you have it

hexdump -x /usr/bin/binaryfile

Answer 4

Edit: Added "bytestream" functionality. If the script name contains the word "stream" (e.g. it's a symlink such as ln -s bash-hexdump bash-hexdump-stream and run as ./bash-hexdump-stream), it will output a continuous stream of hex characters representing the contents of the file. Otherwise its output will look like hexdump -C.

It takes a bunch of trickery since Bash isn't really good at binary:

#!/bin/bash
# bash-hexdump
# by Dennis Williamson - 2010-01-04
# in response to http://stackoverflow.com/questions/2003803/show-hexadecimal-numbers-of-a-file
# usage: bash-hexdump file

if [[ -z "$1" ]]
then
    exec 3<&0                           # read stdin
    [[ -p /dev/stdin ]] || tty="yes"    # no pipe
else
    exec 3<"$1"            # read file
fi

# if the script name contains "stream" then output will be continuous hex digits
# like hexdump -ve '1/1 "%.2x"'
[[ $0 =~ stream ]] && nostream=false || nostream=true

saveIFS="$IFS"
IFS=""                     # disables interpretation of \t, \n and space
saveLANG="$LANG"
LANG=C                     # allows characters > 0x7F
bytecount=0
valcount=0
$nostream && printf "%08x  " $bytecount
while read -s -u 3 -d '' -r -n 1 char    # -d '' allows newlines, -r allows \
do
    ((bytecount++))
    printf -v val "%02x" "'$char"    # see below for the ' trick
    [[ "$tty" == "yes" && "$val" == "04" ]] && break    # exit on ^D
    echo -n "$val"
    $nostream && echo -n " "
    ((valcount++))
    if [[ "$val" < 20 || "$val" > 7e ]]
    then
        string+="."                  # show unprintable characters as a dot
    else
        string+=$char
    fi
    if $nostream && (( bytecount % 8 == 0 ))      # add a space down the middle
    then
        echo -n " "
    fi
    if (( bytecount % 16 == 0 ))   # print 16 values per line
    then
        $nostream && echo "|$string|"
        string=''
        valcount=0
        $nostream && printf "%08x  " $bytecount
    fi
done

if [[ "$string" != "" ]]            # if the last line wasn't full, pad it out
then
    length=${#string}
    if (( length > 7 ))
    then
        ((length--))
    fi
    (( length += (16 - valcount) * 3 + 4))
    $nostream && printf "%${length}s\n" "|$string|"
    $nostream && printf "%08x  " $bytecount
fi
$nostream && echo

LANG="$saveLANG";
IFS="$saveIFS"

The apostrophe trick is documented here. The relevant part says:

If the leading character is a single-quote or double-quote, the value shall be the numeric value in the underlying codeset of the character following the single-quote or double-quote.

Here is some output from the script showing the first few lines of my /bin/bash plus a few more:

00000000  7f 45 4c 46 01 01 01 00  00 00 00 00 00 00 00 00  |.ELF............|
00000010  02 00 03 00 01 00 00 00  e0 1e 06 08 34 00 00 00  |............4...|
00000020  c4 57 0d 00 00 00 00 00  34 00 20 00 09 00 28 00  |.W......4. ...(.|
00000030  1d 00 1c 00 06 00 00 00  34 00 00 00 34 80 04 08  |........4...4...|
. . .
00000150  01 00 00 00 2f 6c 69 62  2f 6c 64 2d 6c 69 6e 75  |..../lib/ld-linu|
00000160  78 2e 73 6f 2e 32 00 00  04 00 00 00 10 00 00 00  |x.so.2..........|
00000170  01 00 00 00 47 4e 55 00  00 00 00 00 02 00 00 00  |....GNU.........|