bash输出星号和引号不正确

Question

so I have this bash function:

function xyz(){
  echo $@
}

now when I run

xyz whaat  "loool * hahhaa"

instead of echoing whaat "loool * hahhaa"

it insteads echoes:

whaat loool all default hahhaa

First of all the quotation marks got stripped off. Secondly the * got replaced by "all default" . This is because the current directory has 2 folders called "all" and "default" hence it think * refers to all directory in the current directory

Is there a way to modify my function so that the output becomes the intended whaat "loool * hahhaa" accordingly (including the quotation marks and *)

I tried doing ${@} "$@" and "${@}" to no avail

Answer 1

The quotation marks are really gone, so don't be expecting anything to put them back. Bash removed them as part of parsing the command line (search for "quote removal" in man bash ).

However, either echo "$@" or echo "${@}" will avoid the glob expansion (replacing * with the directory listing).

You practically never want $@ . Best to get in the habit of writing "$@" .

If you want to see a quoted version of the arguments, you can a bash-extension to printf:

xyz() {
  printf "%q " "$@"
  printf "\n"
}