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Feb 7, 2014 at 4:26 comment added Sophie Swett I finally had the opportunity to test things, and it looks like this answer is incorrect. (Sorry if I'm beating a dead horse here.) If you run seteuid(50) without a matching substitution, overflowuid will not be substituted; instead, the call will fail with EINVAL.
Jan 30, 2014 at 1:23 comment added Sophie Swett I'm not fluent in kernelese, but for what it's worth, I looked at the Linux source code, and it looks like "default to 65534" is only the behavior for mapping a uid from outside a namespace into it. If you're trying to map a uid from inside a namespace out of it, the operation simply fails with EINVAL. So this makes it look like seteuid(50) will fail, since uid 50 as seen from the inside has no mapping outside the namespace.
Jan 30, 2014 at 1:17 comment added Sophie Swett You say your euid will be "still 65534". Since the program is running as uid 1000, won't its euid be 1000 as seen from the outside, and 0 as seen from the inside? If I run seteuid(50), according to you, it sounds like my euid will then be 65534 as seen from the outside. Will my euid be 50 as seen from the inside, as I'd expect?
Jan 21, 2014 at 13:00 history edited slm CC BY-SA 3.0
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Jan 21, 2014 at 12:47 review First posts
Jan 21, 2014 at 12:55
Jan 21, 2014 at 12:30 history answered etherfish CC BY-SA 3.0