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Closures get parent function name

Bash is “kind of function programming language” it dose not have classes. I managed to use encapsulation with Closures, but I want also to do some introspection to find also docker_ parent/super/base function (If you know add comments to define this correctly).

I managed this but with a dirty hack super=${FUNCNAME}. Is there any solution to use kind of PARENT_FUNCNAME? I have such file docker_.sh:

#!/usr/bin/env bash
function docker_ {
    local super=${FUNCNAME}
    function hello {
        echo "INFO" "do ${super}${FUNCNAME}"
    }
    function install {
        echo "INFO" "do ${super}${FUNCNAME}"
        #sudo curl -sSL https://get.docker.com/ | sh || exit 1
    }
    function run {
        echo "INFO" "do ${super}${FUNCNAME}"
        #docker run -d -p 3306:3306 ${DOCKER_IMAGE_NAME} /docker.sh run_mysql
    }
    ${@}
}
${@}

Got some results:

$ ./docker_.sh docker_ hello
INFO do docker_hello

$ ./docker_.sh docker_ run
INFO do docker_run

$ ./docker_.sh docker_ install
INFO do docker_install

Solved

use

${FUNCNAME[1]}
${FUNCNAME[@]:0:${#FUNCNAME[@]}-1} get all list beside main

code:

#!/usr/bin/env bash
function docker_ {
    function hello {
        echo "INFO" "do ${FUNCNAME[1]} ${FUNCNAME}"
    }
    function install {
        echo "INFO" "do > ${FUNCNAME[@]:0:${#FUNCNAME[@]}-1} "
        #sudo curl -sSL https://get.docker.com/ | sh || exit 1
    }
    function run {
        echo "INFO" "do > ${FUNCNAME[@]:0:${#FUNCNAME[@]}-1} "
        #docker run -d -p 3306:3306 ${DOCKER_IMAGE_NAME} /docker.sh run_mysql
    }
    ${@}
}
${@}

Got some results:

    ➜ ./docker_.sh docker_ hello  
    INFO do docker_ hello
    ➜ ./docker_.sh docker_ install
    INFO do > install docker_ 
    ➜ ./docker_.sh docker_ run    
    INFO do > run docker_ 
like image 633
Andrei.Danciuc Avatar asked Oct 19 '25 05:10

Andrei.Danciuc


1 Answers

Bash really isn't a functional programming language. To start with, functions are not first-class objects; you cannot pass a function around. You can pass around the name of a function, but it's just a name; if the name is given a new value, then the old value is lost.

Bash does not have lexical scoping: bash scoping is dynamic. The local command is a command, like any other command. It is not syntactic. If it is not executed, the name is not made local. For example:

f() {
  if [[ $1 == local ]]; then local myvar=local; fi
  myvar=changed
}

$ var=original
$ f local
$ echo $myvar
original
$ f global
$ echo $myvar
changed

And bash does not have closures. You can define a function inside a function, but the function so defined does not carry the scope with it.

$ g() {
>   local myvar=inner
>    f() { echo $myvar; }
>    f
> }
$ myvar=outer
$ g
inner
$ f
outer
$ h() { local myvar=inside_h; f; }
$ h
inside_h
$ f
outer
like image 143
rici Avatar answered Oct 21 '25 02:10

rici



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