[bug] Bash translates >&$var into &>$var for exported functions.

[Dan Douglas]

Hello, In the case of exported functions, Bash interprets a copy descriptor 
followed by an expansion as the >& synonym for &>, resulting in the output 
going to a file named as the value of the FD it's given.  This only applies to 
">&$var" and not "<&$var". I've tested various quoting, Is there some way 
around this?

Gist over here if it's easier to read: https://gist.github.com/1661392

TESTCASE (Overwrites the file named "3" in CWD):
#!/usr/bin/env bash

set -x

f() {
    echo 'hi'
} >&${1}

{ f 3; cat; } <<<'' 3>/dev/stdin

export -f f
export -pf

PS4='* ' BASH_XTRACEFD=4 bash -xc 'f 3; cat' <<<'' 3>/dev/stdin 4>&2

[[ -f 3 ]] && cat ./3
END TESTCASE

OUTPUT:
 ~ $ rm 3; ./exbug
+ f 3
+ echo hi
+ cat
hi
+ export -f f
+ export -pf
f () 
{ 
    echo 'hi'
} &>${1}
declare -fx f
* PS4='* '
* BASH_XTRACEFD=4
+ bash -xc 'f 3; cat'
* f 3
* echo hi
* cat
+ [[ -f 3 ]]
+ cat ./3
hi
END OUTPUT

Bash v. 4.2 w/ patchset 20 on Gentoo Linux amd64.
-- 
Dan Douglas

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