Bash 函数：有没有办法检查变量是否在本地定义？

Question

假设您有一个带有很多选项的函数，并且您决定动态存储它们的值：

#!/bin/bash
fun() {
    local OPTIND __option

    while getopts ":$(printf '%s:' {a..z} {A..Z})" __option
    do
        case $__option in
        [[:alpha:]])
            # ... 
            local "__$__option=$OPTARG"
        esac
    done

    # here you might get a conflict with external variables:
    [[ ${__a:+1} ]] && echo "option a is set"
    [[ ${__b:+1} ]] && echo "option b is set"
    # etc...
    [[ ${__Z:+1} ]] && echo "option Z is set"
}

有没有办法检查变量是否在本地定义？

Answer 1

您绝对可以检测到变量是本地变量（如果它已使用 Bash 的declare, local,语句之一typeset声明）。

这是一个例子：

#!/usr/bin/env bash

a=3
b=2

fn() {
  declare a=1
  b=7
  if local -p a >/dev/null 2>&1; then
    printf 'a is local with value %s\n' "$a"
  else
    printf 'a is not local with value %s\n' "$a"
  fi
  if local -p b >/dev/null 2>&1; then
    printf 'b is local with value %s\n' "$b"
  else
    printf 'b is not local with value %s\n' "$b"
  fi
}


fn

输出是：

a is local with value 1
b is not local with value 7