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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