CODEWITHSUNDEEP
linuxshellash
MOHAMED
MOHAMED

Reputation: 43558

How to evaluate function arguments with for loop

I wan to parse my function arguments with the for loop

func() {
  for arg in $*; do
    echo "$arg"
  cone
}

This works fine if all my arguments are without spaces

func "111" "222" "333"

But it fails for args with spaces

func "111" "222 222" "333"

Upvotes: 0

Views: 52

Answers (2)

choroba
choroba

Reputation: 242103

Don't use $*, use "$@" instead:

for arg in "$@" ; do
     echo "$arg"
done

Explanation: Without double quotes, $* and $@ are identical. Inside double quotes, though, they are different. See "Special parameters" in man bash for details:

* Expands to the positional parameters, starting from one. When the expansion occurs within double quotes, it expands to a single word with the value of each parameter separated by the first character of the IFS special variable. That is, "$*" is equivalent to "$1c$2c...", where c is the first character of the value of the IFS variable. If IFS is unset, the parameters are separated by spaces. If IFS is null, the parameters are joined without intervening separators.

@ Expands to the positional parameters, starting from one. When the expansion occurs within double quotes, each parameter expands to a separate word. That is, "$@" is equivalent to "$1" "$2" ... If the double-quoted expansion occurs within a word, the expansion of the first parameter is joined with the beginning part of the original word, and the expansion of the last parameter is joined with the last part of the original word. When there are no positional parameters, "$@" and $@ expand to nothing (i.e., they are removed).

Upvotes: 2

anubhava
anubhava

Reputation: 785761

Use this function with "$@" instead of unquoted $*:

func() { for arg in "$@"; do echo "$arg"; done; }

Then call it as:

func "111" "222 222" "333"
111
222 222
333

Upvotes: 1

Related Questions