Where does DNF5 take the value for $releasever?

[FrostyX]

I am debugging a weird issue with $releasever not being set on one of my machines (Mock chroot on a bootc system) which lead me into this rabbit hole. When not overriding it with DNF_VAR_MY_VARIABLE=value or any other hack like that, how does DNF5 by itself figure out the value that it sets as $releasever?

I found this code:

dnf5/libdnf5/conf/vars.cpp

Lines 77 to 113 in e9fae8b

	Vars::Impl::Impl(const BaseWeakPtr & base) : base(base) {}
	std::unique_ptr<std::string> Vars::detect_release(const BaseWeakPtr & base, const std::string & install_root_path) {
	std::unique_ptr<std::string> release_ver;
	libdnf5::rpm::RpmLogGuard rpm_log_guard(base);
	auto ts = rpmtsCreate();
	rpmtsSetRootDir(ts, install_root_path.c_str());
	for (auto distroverpkg : DISTROVERPKGS) {
	auto mi = rpmtsInitIterator(ts, RPMTAG_PROVIDENAME, distroverpkg, 0);
	if (auto hdr = rpmdbNextIterator(mi)) {
	const char * version = headerGetString(hdr, RPMTAG_VERSION);
	rpmds ds = rpmdsNew(hdr, RPMTAG_PROVIDENAME, 0);
	while (rpmdsNext(ds) >= 0) {
	if (std::strcmp(rpmdsN(ds), distroverpkg) == 0 && rpmdsFlags(ds) == RPMSENSE_EQUAL) {
	const char * evr = rpmdsEVR(ds);
	if (evr && evr[0] != '\0') {
	version = evr;
	break;
	}
	}
	}
	if (version) {
	// Is the result of rpmdsEVR(ds) valid after rpmdsFree(ds)? Make a copy to be sure.
	release_ver = std::make_unique<std::string>(version);
	}
	rpmdsFree(ds);
	}
	rpmdbFreeIterator(mi);
	if (release_ver) {
	break;
	}
	}
	rpmtsFree(ts);
	return release_ver;
	}

Which I thought makes sense to me, but it doesn't, actually. It goes through this list of packages

dnf5/libdnf5/conf/vars.cpp

Lines 57 to 63 in e9fae8b

	static constexpr const char * DISTROVERPKGS[] = {
	"system-release(releasever)",
	"system-release",
	"distribution-release(releasever)",
	"distribution-release",
	"redhat-release",
	"suse-release"};

and if some of them is installed, it parses some value from it somehow. The strange thing is, none of those packages is installed on any of my systems where $releasever works correctly.

What am I missing, please?

[kontura]

Which I thought makes sense to me, but it doesn't, actually. It goes through this list of packages

Just to be specific this is a list of provides, not concrete packages.
For example on fedora there is fedora-release package which:

dnf rq --provides --installed fedora-release
fedora-release = 43-0.16
fedora-release-variant = 43-0.16
system-release
system-release(43)

The strange thing is, none of those packages is installed on any of my systems where $releasever works correctly.

That is odd, can I ask what system are you running?
I think setup requires fedora-release which I though is pretty much a must have on normal systems.

[FrostyX]

That is odd, can I ask what system are you running?

I have the standard F42 KDE edition:

[jkadlcik@hive ~]$ cat /etc/os-release
NAME="Fedora Linux"
VERSION="42 (KDE Plasma Desktop Edition)"
RELEASE_TYPE=stable
ID=fedora
VERSION_ID=42
...

This returns nothing

[jkadlcik@hive ~]$ dnf rq --provides --installed fedora-release
[jkadlcik@hive ~]$

[kontura]

I see, so then you probably have the fedora-release-kde package variant.

You can check what it provides by:

dnf rq --provides --installed fedora-release-kde

I think it should also provide the fedora-release provide.

If not, you can check which package provides any of the

dnf5/libdnf5/conf/vars.cpp

Lines 57 to 63 in e9fae8b

	static constexpr const char * DISTROVERPKGS[] = {
	"system-release(releasever)",
	"system-release",
	"distribution-release(releasever)",
	"distribution-release",
	"redhat-release",
	"suse-release"};

provides by running for each provide:

dnf rq --whatprovides=system-release --installed

[FrostyX]

Thank you for the explanation @kontura. I forgot to reply, but this helped me understand how it works.