bash脚本中的$@和$*之间的差异

Question

下面有 4 个 bash 片段。我用./script.sh a b c称呼他们

for arg in $@; do 
echo "$arg"
done   ## output "anbnc"
for arg in "$@"; do
echo "$arg"
done  ## output "anbnc" -- I don't know why
for arg in $*; do
echo "$arg"
done  ##    output "anbnc"
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"$@"和"$*"应该是一样的，但事实并非如此。为什么？

Answer 1

如果你有一个脚本foo.sh：

asterisk "$*"
at-sign "$@"

并调用它：

./foo.sh "a a" "b b" "c c"

它相当于：

asterisk "a a b b c c"
at-sign "a a" "b b" "c c"

---

没有引号，它们是相同的：

asterisk $*
at-sign $@

等效于：

asterisk "a" "a" "b" "b" "c" "c"
at-sign "a" "a" "b" "b" "c" "c"

Answer 2

$* 和 $@ 之间的区别是：

"$*" All the positional parameters (as a single word) *

"$@" All the positional parameters (as separate strings)

如果使用 ./my_c $@ 将提供给 bash 脚本的三个命令行参数传递给 C 程序，

你得到的结果ARG[1] == "par1" ARG[2] == "par2" ARG[3] == "par3"

如果使用 .//my_c $* 将提供给 bash 脚本的三个命令行参数传递给 C 程序，

你得到的结果ARG[1] == "par1 par2 par3"

Answer 3

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

如果此脚本按/tmp/testargs.sh a b c 'd e'执行，结果是：

4
1: a
2: b
3: c
4: d e
in quote @, a
in quote @, b
in quote @, c
in quote @, d e
in quote *, a b c d e
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，如果要保留参数的数量，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。