Updated 2018-08-24 14:56:56 by pooryorick

Created by CecilWesterhof.

Getting an UUID on a *NIX System edit

There are some problems with getting the UUID (type 4) on *NIX systems. For one they are not really random.

Because of this I created the following function:
proc getUUIDNix {{secure False}} {
    set fortyeightBits [expr {2 ** 48 - 1}]
    set sixteenBits    [expr {2 ** 16 - 1}]
    set thirtytwoBits  [expr {2 ** 32 - 1}]
    set twelfBits      [expr {2 ** 12 - 1}]

    format %.8x-%.4x-4%.3x-%x%.3x-%.12x \
        [getRandomIntInRangeNix 0 $thirtytwoBits] \
        [getRandomIntInRangeNix 0 $sixteenBits]   \
        [getRandomIntInRangeNix 0 $twelfBits]     \
        [getRandomIntInRangeNix 8 11]               \
        [getRandomIntInRangeNix 0 $twelfBits]     \
        [getRandomIntInRangeNix 0 $fortyeightBits False True]
}

It uses getRandomIntInRangeNix from Random Integers.

A Better way

The above version is a straight implementation. But a more efficient implementation is:
# An UUID is built from 5 hex strings connected by a '-'.
# Their lengths are 8, 4, 4, 4 and 12.
# With version 4 the folowing is necessary:
# - The first digit from the third string is a 4
# - The first digit from the fourth string is 8, 9, A or B.
# The five random hex strings are generated.
# The first digits of the third and fourth strings are changed.
# The UUID is build from the five strings and returned.
# Because binary scan returns lowercase letters toupper is used.
proc getUUIDNix {} {
    binary scan   [getRandomBytesNix 16] H8H4H4H4H12 hex1 hex2 hex3 hex4 hex5
    set    hex3   [string replace $hex3 0 0 4]
    set    oldVal [scan [string index $hex4 0] %x]
    set    newVal [format %X [expr {($oldVal & 3) | 8}]]
    set    hex4   [string replace $hex4 0 0 $newVal]
    string toupper $hex1-$hex2-$hex3-$hex4-$hex5
}

This uses getRandomBytesNix from Get Random Bytes on *NIX.

As always: comments, tips and questions are appreciated.

See also  edit

  • UUID for other implementations

Discussion  edit