How do I pass $@ as a single argument to another command in a bash function?

Question

I have this as a bash function:

function gits() {
  git grep -i -n --color $@ -- $(git rev-parse --show-toplevel);
}

If I run this:

gits def add_crumb

I want:

git grep -i -n --color "def add_crumb" -- $(git rev-parse --show-toplevel)

See [How to iterate over arguments in bash script](http://stackoverflow.com/questions/255898/how-to-iterate-over-arguments-in-bash-script/256225#256225) for an extensive discussion of `$@` vs `$*` vs `"$@"` vs `"$*"`. — Jonathan Leffler, Aug 28 '11 at 19:34

score 10 · Accepted Answer · edited Jul 20 '15 at 11:38

10

The purpose of $@ is specifically to split it into individual arguments. Use "$*" if you don't want that. (Yes, with the double quotes; you should have them in "$@" as well, actually.)

edited Jul 20 '15 at 11:38

fejese

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answered Aug 28 '11 at 18:23

tripleee

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How do I pass $@ as a single argument to another command in a bash function?

1 Answers1