Updated 2018-06-02 15:32:26 by CecilWesterhof

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.