chore: Use geteuid() instead of getuid() to check privilege

[ruihe774]

I believe the correct way is to check geteuid() == 0, not getuid() == 0.

[openshift-ci]

Hi @ruihe774. Thanks for your PR.

I'm waiting for a ostreedev member to verify that this patch is reasonable to test. If it is, they should reply with /ok-to-test on its own line. Until that is done, I will not automatically test new commits in this PR, but the usual testing commands by org members will still work. Regular contributors should join the org to skip this step.

Once the patch is verified, the new status will be reflected by the ok-to-test label.

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[ericcurtin]

This looks correct to me, I tend to favour geteuid over getuid, most of the time that's what people actually want to do. But I'm not sure about every little use-case of OSTree.

What I will say, was there actually a bug or defect found here, etc.?

Or are we just changing for the sake of it because geteuid is the "better" one that's a more true test of rootness?

[ruihe774]

What I will say, was there actually a bug or defect found here, etc.?

Not appeared. But it is easy to produce one. You can set the SUID bit of a program using libostree (e.g. ostree itself). In a program that gains privilege by SUID, getuid() returns the user that invokes the program, and geteuid() returns 0.

[cgwalters]

No one should be including libostree in a setuid binary.
(Really there should be no setuid binaries at all, but that's another thing)

[cgwalters]

/ok-to-test

[ruihe774]

No one should be including libostree in a setuid binary. (Really there should be no setuid binaries at all, but that's another thing)

Yes. However, processes can drop privileges temporarily using seteuid() as well. In this case, getuid() returns 0 and geteuid() returns a unprivileged uid.