bash shell 脚本中的匿名函数
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Anonymous functions in shell scripts
提问by Matty
Is it possible to create something analogous to an anonymous function whose value can be assigned to an array element and later called? I can't seem to find a way to do this in a bash script but perhaps there's a workaround.
是否可以创建类似于匿名函数的东西，其值可以分配给数组元素并在以后调用？我似乎找不到在 bash 脚本中执行此操作的方法，但也许有解决方法。
回答by Ignacio Vazquez-Abrams
Short answer: No.
简短的回答：没有。
Long answer: Nooooooooooooo.
长答案：Nooooooooooooo。
Complete answer: Functions in bash are not first-class objects, therefore there can be no such thing as an anonymous function in bash.
完整答案：bash 中的函数不是一等对象，因此 bash 中不存在匿名函数这样的东西。
回答by Spencer Tipping
It is possible; I wrote a library to do exactly this, though it's a very strange project. The source code is available at http://github.com/spencertipping/bash-lambda. Using this library:
有可能的; 我写了一个库来做到这一点，尽管这是一个非常奇怪的项目。源代码可从http://github.com/spencertipping/bash-lambda 获得。使用这个库：
$ my_array=()
$ my_array[0]=$(fn x 'echo $((x + 1))')
$ my_array[1]=$(fn x 'echo $((x + 2))')
$ ${my_array[0]} 5
6
$ ${my_array[1]} 5
7
$
The trick is to have the fnfunction create a file containing the body of the function, chmod +xthat file, then return its name. This causes stray files to accumulate, which is why the library also implements an asynchronous mark/sweep garbage collector.
诀窍是让fn函数创建一个包含函数主体的chmod +x文件，该文件，然后返回其名称。这会导致杂散文件堆积，这就是该库还实现异步标记/清除垃圾收集器的原因。
回答by choroba
If you really need array to store the functions, you can define named functions and store just their names. You can then call the function as ${array[n]}. Or, you can name them func1.. funcNand then just call func$n.
如果确实需要数组来存储函数，则可以定义命名函数并仅存储它们的名称。然后，您可以将该函数调用为${array[n]}. 或者，您可以将它们命名为func1..funcN然后只需调用func$n.
回答by William Pursell
The common technique is to assign function definitions conditionally:
常用的技术是有条件地分配函数定义：
#!/bin/sh
case in
a) foo() { echo case a; };;
b) foo() { echo case b; };;
*) foo() { echo default; } ;;
esac
foo
回答by edi9999
Create the fn file in your PATH
在您的文件中创建 fn 文件 PATH
#!/bin/sh
printusage () {
printf "Create anonymous function, for example\n"
printf "fn 'echo " "'"
exit 1
}
[ "$#" = "1" ] || printusage
fun=
[ "$fun" = "" ] && printusage
fun_file="$(mktemp /tmp/fun_XXXXXX)"
echo "#!/bin/sh" > "$fun_file"
echo "" >> "$fun_file"
echo "$fun" >> "$fun_file"
chmod u+x "$fun_file"
echo "$fun_file"
You can then do :
然后你可以这样做：
foo=$(fn 'echo ')
${foo} "bar"
回答by slaxor
well bash is turing complete, soo thats perfectly possible ;)
好吧，bash 是图灵完整的，所以这是完全可能的 ;)
but aside from this its not really worth the consideration.
但除此之外，它并不真正值得考虑。
you could simulate such behaviour though with something along this line:
您可以通过以下方式模拟此类行为：
echo myval ; ( foocmd "$_" && barcmd "$_" )
but why?!?
但为什么？！？
