[linux] 쉘 스크립트에서 $ @는 무엇을 의미합니까?

$@는 $*“모든 명령 행 인수”를 의미하는 거의 동일 합니다. 그들은 종종 모든 인수를 다른 프로그램에 전달하는 데 사용됩니다 (따라서 다른 프로그램 주위에 래퍼를 형성 함).

두 구문의 차이점은 공백이 포함 된 인수가 $@있고 큰 따옴표를 넣을 때 나타납니다 .

wrappedProgram "$@"
# ^^^ this is correct and will hand over all arguments in the way
#     we received them, i. e. as several arguments, each of them
#     containing all the spaces and other uglinesses they have.
wrappedProgram "$*"
# ^^^ this will hand over exactly one argument, containing all
#     original arguments, separated by single spaces.
wrappedProgram $*
# ^^^ this will join all arguments by single spaces as well and
#     will then split the string as the shell does on the command
#     line, thus it will split an argument containing spaces into
#     several arguments.

예 : 전화

wrapper "one two    three" four five "six seven"

결과 :

"$@": wrappedProgram "one two    three" four five "six seven"
"$*": wrappedProgram "one two    three four five six seven"
                             ^^^^ These spaces are part of the first
                                  argument and are not changed.
$*:   wrappedProgram one two three four five six seven

답변

for a in "$@"; do something_with "$a"; done

$ cat my-sh
#! /bin/sh
echo "$@"

$ ./my-sh "Hi!"
Hi!

$ put () ( echo "$@" )
$ put "Hi!"
Hi!