I find this function online. It's does creating a directory and changing to directory.
But I want to know every part of it.
function mkdircd () { mkdir -p "$@" && eval cd "\"\$$#\""; }
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2 Answers
You can pass in a list of names. It will create directories for each of them, then cd into the last one.
This does not need eval. I would write it like this:
mkdircd () { mkdir -p "$@" && cd "${!#}"; }
${!#} uses indirect expansion: $# is the number of parameters, so ${!#} is the value of the last parameter
answered May 10, 2016 at 16:09
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Of course if you've used brace expansion it still won't work as expected, will it? E.g.
mkdircd dir{1..3}will put you in the first of the three brace expanded dirs.– WildcardCommented May 10, 2016 at 16:31 -
1Yes, no problem: brace expansion occurs before the function is called, so you are passing 3 distinct arguments in that example. Commented May 10, 2016 at 16:32
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Ahh, good point. Whereas
cd dir{1..3}will put you indir1.– WildcardCommented May 10, 2016 at 16:33 -
1This doesn't need
evalin bash, but usingevalmakes it work in other shells. Commented May 10, 2016 at 22:57
mkdir -p "$@" create all the directories which name are passed as arguments ($@).
The -p option allow to create recursively the directories if they are in directories which don't exist.
eval cd "\"\$$#\"" just go to the last directory: $# give you the number of argument passed, thus \$$# will give you the last argument (i.e: the last directory name you passed as an argument). For example, if there are three arguments, $# is 3, so eval runs the command cd "$3".
The command should actually have been eval cd "\"\${$#}\"". The braces are necessary in many shells when there are 10 arguments are more, because many shells treat something like "$10" as the value of parameter 1 followed by the character 0 and not as the value of parameter 10.
