Hexadecimal number sequence checker in Linux?

Awk could do this easily if the output was in decimal, but it can't parse hexadecimal numbers (at least standard awk can't, some versions such as GNU awk can). You can use bc to do the conversion. This works on all POSIX systems.

{ echo "ibase=16"; cat input.txt; } | bc |
awk 'NR==1 {origin = $0-1}
     $0!=origin+NR {print "Out-of-sequence number at line", NR; exit(1)}' >&2

As said in another answer, it would be nice to do this in awk, but ancient awk (the one probably available in Solaris) doesn't understand hex numbers.

A fast solution is to use wc -l and some ksh math:

#! /bin/ksh

first="0x$(head -n1 infile)"
last="0x$(tail -n1 infile)"
lines=$(wc -l <infile)

if [[ "$(( last - first + 1 - lines ))" -eq 0 ]]; then
    echo "All numbers are in sequence."
else
    echo "Numbers are not in sequence."
fi

Here is one option:

while read x; do echo $((16#$x)); done <yourfile | awk 's && $1!=s+1{exit(1)}{s=$1}'

This shell command will produce an exit status of 1 (on fail) and 0 (on success). This command can be used e.g. in an if-clause like the following to produce the desired output:

if  while read x; do echo $((16#$x)); done < yourfile | awk 's && $1!=s+1{exit(1)}{s=$1}'
then echo All numbers are in sequence
else echo Numbers are not in sequence
fi

(Note that you have you substitute the name "yourfile" by the file name that contains your hex number sequence.)

You can also omit the shell loop and use GNU awk's option -n to process the hex numbers directly:

if  sed 's/^/0x/' <yourfile | awk -n 's && $1+0!=s+1{exit(1)}{s=$1+0}'
then echo All numbers are in sequence
else echo Numbers are not in sequence.
fi

Note: sed is used here to create the syntactically expected format for the hex numbers (with leading 0x).

You can use sort with the option -c to check if the input data is sorted:

-c

Check that the single input file is ordered as specified by the arguments and the collating sequence of the current locale. Output shall not be sent to standard output. The exit code shall indicate whether or not disorder was detected or an error occurred. If disorder (or, with -u, a duplicate key) is detected, a warning message shall be sent to standard error indicating where the disorder or duplicate key was found.

So a simple

LC_ALL=C \
sort -c data 2> /dev/null \
&& echo All numbers are in sequence \
|| echo Numbers are not in sequence

would do.

If you'd like strict sequence it can be easy done by produce etalon sequence and compare it with file

[ $(comm --nocheck-order -3 \
     <(sed '/\S/!d' file) \
     <(printf "%X\n" \
          $(seq $((16#$(sed '/./!d;q' file))) \
                $((16#$(tac file|sed '/./!d;q')))))) ] &&
echo 'Numbers are not in sequence' ||
echo 'All numbers are in sequence'

Just to check if all numbers are hexadecimal:

sed '/^[[:xdigit:] ]*$/!{
        s/.*/Numbers are not in sequence: &/
        q
                        }
     $! d
     s/.*/All numbers are in sequence/' hexadecimal_number.list

• /^[[:xdigit:] ]*$/ check if in line there are hexadecimal symbols and space only (or nothing - empty line)
• ! reverse match, so execute commands in {} just if there some other symbols than described above
• s/.*/.../;q substitute all line by Numbers are not in sequence than quit script with printing line
• $! d if script didn't execute previous (pattern match) and if line is not last delete it it start with next line
• s/.*/.../ if script reach last line (so all lines match pattern) substitute last line by All numbers are in sequence an print it

Given the input file hexlist.txt, this works in ksh and bash. It uses the numinterval util. If numinterval isn't available, uncomment the faker function:

# fake `numinterval` function
# numinterval() { printf '%s %s r - p\n' $(sed '1!{$!p}') | dc ; }

n=; eval printf '%s\\n' $(printf '$((16#%s\n)) ' $(<hexlist.txt)) | 
numinterval | grep -qw 1 || n=not ; echo All numbers are $n in sequence