Linux shell 脚本函数返回一个字符串
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shell script function return a string
提问by Lolly
I am new to shell scripts, I am trying to create a simple function which will return the concatenated two strings that are passed as parameters. I tried with below code
我是 shell 脚本的新手，我正在尝试创建一个简单的函数，该函数将返回作为参数传递的连接的两个字符串。我试过下面的代码
function getConcatenatedString() {
echo "String1 "
echo "String2 "
str=/
echo "Concatenated String ${str}"
echo "${str}"
}
//I am calling the above function
//我正在调用上面的函数
constr=$(getConcatenatedString "hello" "world")
echo "printing result"
echo "${constr}"
echo "exit"
I see the below output when running the script with above code,
使用上述代码运行脚本时，我看到以下输出，
printing result
String1 hello
String2 world
Concatenated String hello/world
hello/world
exit
If you look at the code I am first calling the function and then I am echoing "printing result" statement, but the result is first comes the "printing result" and echos the statement inside the function. Is the below statement calling the function
如果您查看代码，我首先调用函数，然后回显“打印结果”语句，但结果首先是“打印结果”并回显函数内部的语句。下面的语句是调用函数吗
constr=$(getConcatenatedString "hello" "world")
or
或者
echo ${constr}
is calling the function ?
正在调用函数？
Because if I comment out #echo ${constr} then nothing is getting echoed !!! Please clarify me.
因为如果我注释掉 #echo ${constr} 那么什么都不会得到回应！！！请澄清我。
采纳答案by paxdiablo
The first is calling the function and storing allof the output (four echostatements) into $constr.
第一个是调用函数并将所有输出（四个echo语句）存储到$constr.
Then, after return, you echo the preamble printing result, $constr(consisting of four lines) and the exit message.
然后，返回后，您回显 preamble printing result, $constr（由四行组成）和退出消息。
That's how $()works, it captures the entire standard output from the enclosed command.
这就是$()工作原理，它从包含的命令中捕获整个标准输出。
It sounds like you want to see some of the echostatements on the console rather than capturing them with the $(). I think you should just be able to send them to standard error for that:
听起来您想echo在控制台上查看某些语句，而不是使用$(). 我认为你应该能够将它们发送到标准错误：
echo "String1 " >&2
回答by William Pursell
paxdiablo's solution is correct. You cannot return a string from a function, but you can capture the output of the function or return an integer value that can be retrieved by the caller from $?. However, since all shell variables are global, you can simply do:
paxdiablo 的解决方案是正确的。您不能从函数返回字符串，但可以捕获函数的输出或返回可由调用者从 中检索的整数值$?。但是，由于所有 shell 变量都是全局变量，您可以简单地执行以下操作：
getConcatenatedString() { str="/"; }
getConcatenatedString hello world
echo "Concatenated String ${str}"
Note that the functionkeyword is redundant with (), but functionis less portable.
请注意，function关键字与 是多余的()，但function可移植性较差。
回答by mr.spuratic
A more flexible, but slightly harder to understand approach is to pass a variable name, and use evalso that the variable becomes set in the caller's context (either a global or a function local). In bash:
一种更灵活但稍微难以理解的方法是传递一个变量名，并使用它eval以便在调用者的上下文中设置变量（全局或局部函数）。在 bash 中：
function mylist()
{
local _varname= _p _t
shift
for _p in "$@"; do
_t=$_t[$_p]
done
eval "$_varname=$_t"
}
mylist tmpvar a b c
echo "result: $tmpvar"
On my Linux desktop (bash-3.2) it's approx 3-5x faster (10,000 iterations) than using ``, since the latter has process creation overheads.
在我的 Linux 桌面 (bash-3.2) 上，它比使用 快约 3-5 倍（10,000 次迭代）``，因为后者具有进程创建开销。
If you have bash-4.2, its declare -gallows a function to set a global variable, so you can replace the unpretty evalwith:
如果你有 bash-4.2，它declare -g允许一个函数设置一个全局变量，所以你可以eval用：
declare -g $_varname="$_t"
The evalmethod is similar to TCL'supvar 1, and declare -gis similar to upvar #0.
该eval方法类似于TCL的upvar 1，和declare -g类似upvar #0。
Some shell builtins support something similar, like bash's printfwith "-v", again saving process creation by assigning directly to a variable instead of capturing output (~20-25x faster for me).
一些 shell 内置函数支持类似的东西，比如printf带有“-v”的bash ，再次通过直接分配给变量而不是捕获输出来保存进程创建（对我来说快 20-25 倍）。
