A recent vulnerability, CVE-2014-6271, in how Bash interprets environment variables was disclosed. The exploit relies on Bash parsing some environment variable declarations as function definitions, but then continuing to execute code following the definition:
$ x='() { echo i do nothing; }; echo vulnerable' bash -c ':'
vulnerable
But I don't get it. There's nothing I've been able to find in the Bash manual about interpreting environment variables as functions at all (except for inheriting functions, which is different). Indeed, a proper named function definition is just treated as a value:
$ x='y() { :; }' bash -c 'echo $x'
y() { :; }
But a corrupt one prints nothing:
$ x='() { :; }' bash -c 'echo $x'
$ # Nothing but newline
The corrupt function is unnamed, and so I can't just call it. Is this vulnerability a pure implementation bug, or is there an intended feature here, that I just can't see?
Update
Per Barmar's comment, I hypothesized the name of the function was the parameter name:
$ n='() { echo wat; }' bash -c 'n'
wat
Which I could swear I tried before, but I guess I didn't try hard enough. It's repeatable now. Here's a little more testing:
$ env n='() { echo wat; }; echo vuln' bash -c 'n'
vuln
wat
$ env n='() { echo wat; }; echo $1' bash -c 'n 2' 3 -- 4
wat
…so apparently the args are not set at the time the exploit executes.
Anyway, the basic answer to my question is, yes, this is how Bash implements inherited functions.
See Question&Answers more detail:
os 与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…