下面有4个bash片段。 我用./script.sh abc调用它们
for arg in $@; do echo "$arg" done ## output "a\nb\nc" for arg in "$@"; do echo "$arg" done ## output "a\nb\nc" -- I don't know why for arg in $*; do echo "$arg" done ## output "a\nb\nc" for arg in "$*"; do echo "$arg" done ## output "abc"
我不知道$@和$*之间的确切区别是什么，
我认为"$@"和"$*"应该是相同的，但它们不是。 为什么？
如果你有一个脚本foo.sh ：
asterisk "$*" at-sign "$@"
并用以下方式调用它：
./foo.sh "aa" "bb" "cc"
它相当于：
asterisk "aabbcc" at-sign "aa" "bb" "cc"
没有引号，他们是一样的：
asterisk $* at-sign $@
将相当于：
asterisk "a" "a" "b" "b" "c" "c" at-sign "a" "a" "b" "b" "c" "c"
$ *和$ @之间的区别是::
"$*" All the positional parameters (as a single word) *
"$@" All the positional parameters (as separate strings)
如果您使用./my_c $ @传递给bash脚本的三个命令行参数到一个C程序，
你会得到结果ARG[1] == "par1" ARG[2] == "par2" ARG[3] == "par3"
如果您使用./my_c $ *将三个命令行参数传递给C程序，
你得到结果ARG[1] == "par1 par2 par3"
这在shell脚本中很重要：例如脚本testargs.sh
#! /bin/bash -p echo $# for i in $(seq 1 $#) do echo "$i: ${!i}" done for val in "$@"; do echo "in quote @, $val" done for val in "$*"; do echo "in quote *, $val" done for val in $@; do echo "not in quote @, $val" done for val in $*; do echo "not in quote *, $val" done
如果此脚本以/tmp/testargs.sh abc 'd e' ，则结果为：
4 1: a 2: b 3: c 4: de in quote @, a in quote @, b in quote @, c in quote @, de in quote *, abcde not in quote @, a not in quote @, b not in quote @, c not in quote @, d not in quote @, e not in quote *, a not in quote *, b not in quote *, c not in quote *, d not in quote *, e
因此，如果要保存参数的数量，请始终使用“$ @”或使用for i in $(seq 1 $#)循环遍历每个参数。 没有引号，两者都是一样的。