Random number generation in Solaris?

Question

What is the best way to generate random numbers in Solaris?

I can not seem to find a good answer for this. Most results do not work in my environment. There is a variable or command RAND that seems logical it would work in some manner similar to $RANDOM which I see in most of my searches but it always produces 0.

I have found this command

od -X -A n /dev/random | head -2

Which seems very random but the return format is odd (to me).

     140774 147722 131645 061031 125411 053337 011722 165106
     066120 073123 040613 143651 040740 056675 061051 015211

Currently using:

-bash-3.2$ uname -a
SunOS XXXXXXXXX 5.10 Generic_150400-29 sun4v sparc SUNW,SPARC-Enterprise-T5120

Answer 1

$RANDOM is available in ksh and in bash, but not in /bin/sh. The value is a random number between 0 and 32768, and is not suitable for cryptographic use.

Reading from /dev/random generates a stream of random bytes which is suitable for cryptographic use. Since these are arbitrary bytes, potentially including null bytes, you can't store them in a shell variable. You can store $n bytes in a file with

</dev/random dd ibs=1 count=$n >rnd

You can use od to transform these bytes into a printable representation using octal or hexadecimal values. If you find the output “strange”, well, maybe you should pick different od options.

Another option to obtain a printable representation is to call uuencode to produce Base64:

</dev/random dd ibs=1 count=$n | uuencode -m _ | sed -e '1d' -e '$d'

Answer 2

This are three ways to get n bytes, expressed as 2*n Hexadecimal digits:

#!/bin/bash
n=64
# Read n bytes from urandom (in hex).
xxd -l "$n" -p                    /dev/urandom | tr -d " \n" ; echo
od  -vN "$n" -An -tx1             /dev/urandom | tr -d " \n" ; echo
hexdump -vn "$n" -e ' /1 "%02x"'  /dev/urandom ; echo

Reading from urandom (and Please, do use urandom).

The hex digits may be redirected to a file or stored in a variable:

 a="$(xxd -l "$n" -p /dev/urandom)"

And you could get the original bytes using xxd, as simple as:

## Warning, this produces binary values

echo "$a" | xxd -r -p      # If the Hex digits are in a variable.
xxd -r -p "$rndfile"       # If the Hex digits are in a file.

If xxd is not available, and assuming the hex digits are in $a.
You could use this bash code:

#!/bin/bash

>"$rndfile"                            # erase the file

for (( i=0; i<${#a}/2; i++ )); do      # do a loop over all byte values.

    # convert 2 hex digits to the byte unsigned value:
    printf '%b' $(printf '\\0%o' "0x${a:$i*2:2}") >> "$rndfile"

    b="${b#??}"  # chop out two hexadecimal digits.

done

Or you may try this sh compatible code (should run in bash comfortably). Yes it should be sh compatible. Report if you find problems.

Only tested in dash (but should run in some others).

#!/bin/sh

i=$((${#a}/2))          # The (length of $a ) / 2 is the number of bytes.
b="$a"                  # use a temporal variable. It will be erased.
: >"$rndfile"           # erase file contents

while [ $i != 0 ]; do

    # Write one byte transformed to binary:
    printf '%b' $(printf '\\0%o' "0x${b%"${b#??}"}") >> "$rndfile"

    b="${b#??}"  # chop out two hexadecimal digits.
    i=$((i-1))   # One byte less to finish.

done

Answer 3

</dev/random \
dd bs="${num_len}" count=1 | LC_ALL=C \
tr -c 0-9 "$(printf '[%d*24]' 9 8 7 6 5 4 3 2 1)[0*]"

that ought to work well for base 10 and wastes nothing, though the digit distribution is very slightly favored for 0.

with Solaris v<11 you'll want to use ` backticks rather than $(...) for the command substitution.