为了说明我对Andy的回答的评论，使用额外的文件描述符操作来避免使用/dev/tty:

#!/bin/bash

exec 3>&1

returnString() {
    exec 4>&1 >&3
    local s=$1
    s=${s:="some default string"}
    echo "writing to stdout"
    echo "writing to stderr" >&2
    exec >&4-
    echo "$s"
}

my_string=$(returnString "$*")
echo "my_string:  [$my_string]"

不过还是很恶心。

以上所有答案都忽略bash手册页中所述内容。

函数中声明的所有变量都将与调用环境共享。 所有在本地声明的变量将不会被共享。

示例代码

#!/bin/bash

f()
{
    echo function starts
    local WillNotExists="It still does!"
    DoesNotExists="It still does!"
    echo function ends
}

echo $DoesNotExists #Should print empty line
echo $WillNotExists #Should print empty line
f                   #Call the function
echo $DoesNotExists #Should print It still does!
echo $WillNotExists #Should print empty line

和输出

$ sh -x ./x.sh
+ echo

+ echo

+ f
+ echo function starts 
function starts
+ local 'WillNotExists=It still does!'
+ DoesNotExists='It still does!'
+ echo function ends 
function ends
+ echo It still 'does!' 
It still does!
+ echo

同样在pdksh和ksh下，这个脚本也做同样的事情!

您还可以捕获函数输出:

#!/bin/bash
function getSomeString() {
     echo "tadaa!"
}

return_var=$(getSomeString)
echo $return_var
# Alternative syntax:
return_var=`getSomeString`
echo $return_var

看起来很奇怪，但比使用全局变量更好。传递参数和往常一样，只是把它们放在大括号或反勾号内。

Bash自2014年2月4.3版(?)起，除了“eval”之外，还明确支持引用变量或名称引用(namerefs)，具有相同的性能和间接效果，并且在你的脚本中可能更清晰，也更难“忘记'eval'而不得不修复此错误”:

declare [-aAfFgilnrtux] [-p] [name[=value] ...]
typeset [-aAfFgilnrtux] [-p] [name[=value] ...]
  Declare variables and/or give them attributes
  ...
  -n Give each name the nameref attribute, making it a name reference
     to another variable.  That other variable is defined by the value
     of name.  All references and assignments to name, except for⋅
     changing the -n attribute itself, are performed on the variable
     referenced by name's value.  The -n attribute cannot be applied to
     array variables.
...
When used in a function, declare and typeset make each name local,
as with the local command, unless the -g option is supplied...

还有:

PARAMETERS A variable can be assigned the nameref attribute using the -n option to the declare or local builtin commands (see the descriptions of declare and local below) to create a nameref, or a reference to another variable. This allows variables to be manipulated indirectly. Whenever the nameref variable is⋅ referenced or assigned to, the operation is actually performed on the variable specified by the nameref variable's value. A nameref is commonly used within shell functions to refer to a variable whose name is passed as an argument to⋅ the function. For instance, if a variable name is passed to a shell function as its first argument, running declare -n ref=$1 inside the function creates a nameref variable ref whose value is the variable name passed as the first argument. References and assignments to ref are treated as references and assignments to the variable whose name was passed as⋅ $1. If the control variable in a for loop has the nameref attribute, the list of words can be a list of shell variables, and a name reference will be⋅ established for each word in the list, in turn, when the loop is executed. Array variables cannot be given the -n attribute. However, nameref variables can reference array variables and subscripted array variables. Namerefs can be⋅ unset using the -n option to the unset builtin. Otherwise, if unset is executed with the name of a nameref variable as an argument, the variable referenced by⋅ the nameref variable will be unset.

例如(EDIT 2:(谢谢你Ron)在函数内部变量名的命名空间(前缀)，以最小化外部变量冲突，这最终应该正确地回答了Karsten在评论中提出的问题):

# $1 : string; your variable to contain the return value
function return_a_string () {
    declare -n ret=$1
    local MYLIB_return_a_string_message="The date is "
    MYLIB_return_a_string_message+=$(date)
    ret=$MYLIB_return_a_string_message
}

测试这个例子:

$ return_a_string result; echo $result
The date is 20160817

请注意，bash“declare”内置在函数中使用时，默认情况下会使声明的变量为“local”，并且“-n”也可以与“local”一起使用。

我更喜欢区分“重要的声明”变量和“无聊的本地”变量，因此以这种方式使用“声明”和“本地”作为文档。

编辑1 -(对Karsten下面的评论的回应)-我不能再在下面添加评论了，但Karsten的评论让我思考，所以我做了以下测试，工作良好，AFAICT - Karsten如果你读了这篇文章，请从命令行提供一组准确的测试步骤，显示你假设存在的问题，因为以下步骤工作得很好:

$ return_a_string ret; echo $ret
The date is 20170104

(我刚刚将上面的函数粘贴到bash术语中后运行了这个程序——正如您所看到的，结果运行得很好。)

#实现一个通用的函数返回堆栈:

STACK=()
push() {
  STACK+=( "${1}" )
}
pop() {
  export $1="${STACK[${#STACK[@]}-1]}"
  unset 'STACK[${#STACK[@]}-1]';
}

#用法:

my_func() {
  push "Hello world!"
  push "Hello world2!"
}
my_func ; pop MESSAGE2 ; pop MESSAGE1
echo ${MESSAGE1} ${MESSAGE2}

我想，所有的选择都已经列举出来了。选择一种可以归结为最适合您的特定应用程序的样式，因此，我想提供一种我认为有用的特定样式。在bash中，变量和函数不在同一个命名空间中。因此，将同名变量视为函数值是一种约定，如果严格应用它，我发现这种约定可以最大限度地减少名称冲突并增强可读性。一个来自现实生活的例子:

UnGetChar=
function GetChar() {
    # assume failure
    GetChar=
    # if someone previously "ungot" a char
    if ! [ -z "$UnGetChar" ]; then
        GetChar="$UnGetChar"
        UnGetChar=
        return 0               # success
    # else, if not at EOF
    elif IFS= read -N1 GetChar ; then
        return 0           # success
    else
        return 1           # EOF
    fi
}

function UnGetChar(){
    UnGetChar="$1"
}

下面是一个使用这些函数的例子:

function GetToken() {
    # assume failure
    GetToken=
    # if at end of file
    if ! GetChar; then
        return 1              # EOF
    # if start of comment
    elif [[ "$GetChar" == "#" ]]; then
        while [[ "$GetChar" != $'\n' ]]; do
            GetToken+="$GetChar"
            GetChar
        done
        UnGetChar "$GetChar"
    # if start of quoted string
    elif [ "$GetChar" == '"' ]; then
# ... et cetera

如您所见，返回状态是供您在需要时使用的，如果不需要则忽略它。“返回”变量同样可以被使用或忽略，但当然只有在函数被调用之后。

当然，这只是一种惯例。您可以在返回之前不设置相关值(因此我的约定总是在函数开始时将其为空)，或者通过再次调用函数(可能是间接地)来破坏它的值。不过，如果我发现自己大量使用bash函数，我觉得这种约定非常有用。

相反的情绪，这是一个迹象，一个应该。“转到perl”，我的哲学是，对于管理任何语言的复杂性，约定总是很重要的。