bash `case` 语句将输入分类为非整数和整数

摘要：我想使用一个bash case语句（在其他代码中）对输入进行分类，以确定它们是否是

• 一个正整数
• 一个负整数
• 零
• 一个空字符串
• 一个非整数字符串

可执行代码如下，正确分类以下输入：

• ''
• word
• a\nmultiline\nstring
• 2.1
• -3

但是将以下两者都归类为...负整数:-(

• 0
• 42

细节：

将以下内容保存到文件（例如/tmp/integer_case_statement.sh）中chmod，然后运行它：

#!/usr/bin/env bash

### This should be a simple `bash` `case` statement to classify inputs as
### {positive|negative|zero|non-} integers.
### Trying extglob, since my previous integer-match patterns failed.
### Gotta turn that on before *function definition* per https://stackoverflow.com/a/34913651/915044
shopt -s extglob

declare cmd=''

function identify() {
    local -r it=${1}  # no quotes in case it's multiline
#    shopt -s extglob # can't do that here
    case ${it} in
        '')
            # empty string, no need for `if [[ -z ...`
            >&2 echo 'ERROR: null arg'
            ;;
        ?(-|+)+([[:digit:]]))
            # it's an integer, so just say so and fallthrough
            >&2 echo 'DEBUG: int(it), fallthrough'
            ;&
        -+([[:digit:]]))
            # it's negative: just return it
            >&2 echo 'DEBUG: int(it) && (it < 0), returning it'
            echo "${it}"
            ;;
        0)
            # it's zero: that's OK
            >&2 echo 'DEBUG: int(it) && (it == 0), returning it'
            echo '0'
            ;;
        ++([[:digit:]]))
            # it's positive: just return it
            >&2 echo 'DEBUG: int(it) && (it > 0), returning it'
            echo "${it}"
            ;;
        *)
            # not an integer, just return it
            >&2 echo 'DEBUG: !int(it)'
            echo "${it}"
            ;;
    esac
} # end function identify

echo -e "'bash --version'==${BASH_VERSION}\n"

echo "identify '':"
identify ''
echo
# > ERROR: null arg

echo 'identify word:'
identify word
echo
# > DEBUG: !int(it)
# > word

echo 'identify a
multiline
string:'
identify 'a
multiline
string'
echo
# > DEBUG: !int(it)
# > a
# > multiline
# > string

echo 'identify 2.1:'
identify 2.1
echo
# > DEBUG: !int(it)
# > 2.1

echo 'identify -3:'
identify -3
echo
# > DEBUG: int(it), fallthrough
# > DEBUG: int(it) && (it < 0), returning it
# > -3

echo 'identify 0:'
identify 0
echo
# > DEBUG: int(it), fallthrough
# > DEBUG: int(it) && (it < 0), returning it
# > 0

echo 'identify 42:'
identify 42
echo
# > DEBUG: int(it), fallthrough
# > DEBUG: int(it) && (it < 0), returning it
# > 42

exit 0

当前输出被内联在文件中，但为了便于阅读，这里单独列出我当前的输出：

'bash --version'==4.3.30(1)-release

identify '':
ERROR: null arg

identify word:
DEBUG: !int(it)
word

identify a
multiline
string:
DEBUG: !int(it)
a
multiline
string

identify 2.1:
DEBUG: !int(it)
2.1

identify -3:
DEBUG: int(it), fallthrough
DEBUG: int(it) && (it < 0), returning it
-3

identify 0:
DEBUG: int(it), fallthrough
DEBUG: int(it) && (it < 0), returning it
0

identify 42:
DEBUG: int(it), fallthrough
DEBUG: int(it) && (it < 0), returning it
42

后两个输入是我的问题：为什么 case 语句识别

• 0 作为负整数（而不是 0）
• 42 作为负整数（而不是正整数）

? 感谢您的帮助。