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I was recently tripped up by a bug in Puppet, PUP-1244, dealing with how it compares package versions. All of Puppet’s Package types assumed semantic versioning, and that’s far from the case for RPMs and therefore Yum. This manifested itself in how Puppet validates package installations - if a version was explicitly specified and Yum/RPM could install it, puppet would shell out to them and install the package, but then report a failure in its post-install validation, as the exact version string specified isn’t present. For example, many RedHat/CentOS packages (such as those from EPEL) include a release string with the major version of the distribution they were packged for - i.e. “.el5” or “.el6”. If Puppet was instructed to install package “foo” version “1.2.3”, but the actual package in the repositories was “foo-1.2.3-el5”, Puppet would cause the package to be installed, but then report failure.

I cut a pull request against the Puppet4 branch to fix these issues, essentially re-implementing yum and rpm’s version comparison logic in Ruby. It took me a few days of research and sorting through source code (and in the process I found that yum, despite its use in so many distributions, has no unit tests at all) but I finally got it finished. In the process, I found out exactly how… weird… RPM’s version comparison rules are.

Package Naming and Parsing

RPM package names are made up of five parts; the package name, epoch, version, release, and architecture. This format is commonly referred to as the acronym NEVRA. The epoch is not always included; it is assumed to be zero (0) on any packages that lack it explicitly. The format for the whole string is n-e:v-r.a. For my purposes, I was only really concerned with comparing the EVR portion; Puppet knows about package names and the bug herein was with what Puppet calls the “version” (EVR in yum/rpm parlance). Parsing is pretty simple:

  • If there is a : in the string, everything before it is the epoch. If not, the epoch is zero.
  • If there is a - in the remaining string, everything before the first - is the version, and everything after it is the release. If there isn’t one, the release is considered null/nill/None/whatever.

How Yum Compares EVR

Once the package string is parsed into its EVR components, yum calls rpmUtils.miscutils.compareEVR(), which does some data type massaging for the inputs, and then calls out to rpm.labelCompare() (found in rpm.git/python/header-py.c). labelCompare() sets each epoch to “0” if it was null/Nonem, and then uses compare_values() to compare each EVR portion, which in turn falls through to a function called rpmvercmp() (see below). The algorithm for labelCompare() is as follows:

  1. Set each epoch value to 0 if it’s null/None.
  2. Compare the epoch values using compare_values(). If they’re not equal, return that result, else move on to the next portion (version). The logic within compare_values() is that if one is empty/null and the other is not, the non-empty one is greater, and that ends the comparison. If neither of them is empty/not present, compare them using rpmvercmp() and follow the same logic; if one is “greater” (newer) than the other, that’s the end result of the comparison. Otherwise, move on to the next component (version).
  3. Compare the versions using the same logic.
  4. Compare the releases using the same logic.
  5. If all of the components are “equal”, the packages are the same.

The real magic, obviously, happens in rpmvercmp(), the rpm library function to compare two versions (or epochs, or releases). That’s also where the madness happens.

How RPM Compares Version Parts

RPM is written in C. Converting all of the buffer and pointer processing for these strings over to Ruby was quite a pain. That being said, I didn’t make this up, this is actually the algorithm that rpmvercmp() (lib/rpmvercmp.c) uses to compare version “parts” (epoch, version, release). This function returns 0 if the strings are equal, 1 if a (the first string argument) is newer than b (the second string argument), or -1 if a is older than b. Also keep in mind that this uses pointers in C, so it works by removing a sequence of 0 or more characters from the front of each string, comparing them, and then repeating for the remaining characters in each string until something is unequal, or a string reaches its end.

  1. If the strings are binary equal (a == b), they’re equal, return 0.
  2. Loop over the strings, left-to-right.
    1. Trim anything that’s not [A-Za-z0-9] or tilde (~) from the front of both strings.
    2. If both strings start with a tilde, discard it and move on to the next character.
    3. If string a starts with a tilde and string b does not, return -1 (string a is older); and the inverse if string b starts with a tilde and string a does not.
    4. End the loop if either string has reached zero length.
    5. If the first character of a is a digit, pop the leading chunk of continuous digits from each string (which may be ” for b if only one a starts with digits). If a begins with a letter, do the same for leading letters.
    6. If the segement from b had 0 length, return 1 if the segment from a was numeric, or -1 if it was alphabetic. The logical result of this is that if a begins with numbers and b does not, a is newer (return 1). If a begins with letters and b does not, then a is older (return -1). If the leading character(s) from a and b were both numbers or both letters, continue on.
    7. If the leading segments were both numeric, discard any leading zeros and whichever one is longer wins. If a is longer than b (without leading zeroes), return 1, and vice-versa. If they’re of the same length, continue on.
    8. Compare the leading segments with strcmp() (or <=> in Ruby). If that returns a non-zero value, then return that value. Else continue to the next iteration of the loop.
  3. If the loop ended (nothing has been returned yet, either both strings are totally the same or they’re the same up to the end of one of them, like with “1.2.3” and “1.2.3b”), then the longest wins - if what’s left of a is longer than what’s left of b, return 1. Vice-versa for if what’s left of b is longer than what’s left of a. And finally, if what’s left of them is the same length, return 0.

Well there you have it. Quite convoluted. And full of things like the “~” magic character (“~1” is always older than “9999zzzz”).


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