$*和$@的不一致的行为

#!/bin/bash

#  内部Bash变量"$*"和"$@"的古怪行为,
#  都依赖于它们是否被双引号引用起来.
#  单词拆分与换行的不一致的处理.


set -- "First one" "second" "third:one" "" "Fifth: :one"
# 设置这个脚本的参数, $1, $2, 等等.

echo

echo 'IFS unchanged, using "$*"'
c=0
for i in "$*"               # 引用起来
do echo "$((c+=1)): [$i]"   # 这行在下边每个例子中都一样.
                            # 打印参数.
done
echo ---

echo 'IFS unchanged, using $*'
c=0
for i in $*                 # 未引用
do echo "$((c+=1)): [$i]"
done
echo ---

echo 'IFS unchanged, using "$@"'
c=0
for i in "$@"
do echo "$((c+=1)): [$i]"
done
echo ---

echo 'IFS unchanged, using $@'
c=0
for i in $@
do echo "$((c+=1)): [$i]"
done
echo ---

IFS=:
echo 'IFS=":", using "$*"'
c=0
for i in "$*"
do echo "$((c+=1)): [$i]"
done
echo ---

echo 'IFS=":", using $*'
c=0
for i in $*
do echo "$((c+=1)): [$i]"
done
echo ---

var=$*
echo 'IFS=":", using "$var" (var=$*)'
c=0
for i in "$var"
do echo "$((c+=1)): [$i]"
done
echo ---

echo 'IFS=":", using $var (var=$*)'
c=0
for i in $var
do echo "$((c+=1)): [$i]"
done
echo ---

var="$*"
echo 'IFS=":", using $var (var="$*")'
c=0
for i in $var
do echo "$((c+=1)): [$i]"
done
echo ---

echo 'IFS=":", using "$var" (var="$*")'
c=0
for i in "$var"
do echo "$((c+=1)): [$i]"
done
echo ---

echo 'IFS=":", using "$@"'
c=0
for i in "$@"
do echo "$((c+=1)): [$i]"
done
echo ---

echo 'IFS=":", using $@'
c=0
for i in $@
do echo "$((c+=1)): [$i]"
done
echo ---

var=$@
echo 'IFS=":", using $var (var=$@)'
c=0
for i in $var
do echo "$((c+=1)): [$i]"
done
echo ---

echo 'IFS=":", using "$var" (var=$@)'
c=0
for i in "$var"
do echo "$((c+=1)): [$i]"
done
echo ---

var="$@"
echo 'IFS=":", using "$var" (var="$@")'
c=0
for i in "$var"
do echo "$((c+=1)): [$i]"
done
echo ---

echo 'IFS=":", using $var (var="$@")'
c=0
for i in $var
do echo "$((c+=1)): [$i]"
done

echo

# 使用ksh或者zsh -y来试试这个脚本.

exit 0