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.