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lang-bash
$BASH_ARGVdoesn't work inside a bash function (unless I'm doing something wrong).BASH_ARGVwill yield you is the value that the last arg that was given was, instead of simply "the last value". For example!: if you provide one single argument, then you call shift,${@:$#}will produce nothing (because you shifted out the one and only argument!), however,BASH_ARGVwill still give you that (formerly) last argument.${!#}get nothing when execute script withsh XXX.sh 1 2 3/bin/bash). You ran it through POSIX sh (/bin/sh), which on many systems is not bash.