Clone a variable in bash?

Question

Before explaining my bash problem let me give you some context:

I am writing some scripts using a bash "framework" we use at my current job. One of the feature of the framework is to init a sets of environments variables useful to run jobs on our cluster infrastructure.

These variables depend on a date specified by $YY, $mm and $dd which are also environment variables (yes, this is wired). To use the framework you start by defining the date and then you call a function to init other vars. This works fine when you write scripts that need variables for a specific day only. Today I am writing something which needs variables for 2 different days. Writing this I face a strange issue. For you to better understand the problem I wrote this code that simulate it:

#!/bin/bash

function assign(){
    date=$1
    date[1]=$YY
    date[2]=$mm
    date[3]=$dd
}

function display() {
    date=$1
    echo "${date[1]}/${date[2]}/${date[3]}"
}

export YY=2012
export mm=09
export dd=20
declare -a my_date1=()
assign $my_date1

export YY=2012
export mm=08
export dd=20
declare -a my_date2=()
assign $my_date2

display $my_date1
display $my_date2

The expected output is:

2012/09/20
2012/08/20

But the output is:

2012/08/20
2012/08/20

At first I thought that the assign function filled the array with reference to $YY, $mm and $dd instead of their values. But then I try with the following code and it doesn't change the result.

date[1]=$(echo $YY)
date[2]=$(echo $mm)
date[3]=$(echo $dd)

Can somebody explain me what append? Maybe something wired with date=$1...

Try changing the shebang line to #!/bin/bash -xv to see how bash interprets the code. — choroba
– choroba, Commented Sep 21, 2012 at 19:12
There is never a need to export a variable twice, and there is no apparent need in this script to export them even once. — William Pursell
– William Pursell, Commented Sep 21, 2012 at 20:01
@WilliamPursell: it's why I put some context to my question! I did these things because of the crazy framework we use at my current job. I asked this question to understand this behavior not to solve the problem I faced at work. Even if this script is stupid I learned something from understanding it. Thanks to chepner for his answer and choroba for his suggestion, it enable me to fully understand the answer. — a.b.d
– a.b.d, Commented Sep 21, 2012 at 21:32

chepner · Accepted Answer · 2012-09-21 20:22:33Z

Arrays are not passed by either value or reference in bash. Rather, the value of the expansion of the array is passed by value. When you write

assign $my_date1

the date variable inside assign is null, since $my_date1 expands to an empty string and disappears after word-splitting before the function is called. As a result, $1 is unset.

But date, being a global variable because it was not declared as local, is set correctly using YY et al, then reset on the second call to assign.

Also, note that the first line of your functions does not make date a reference to the argument; it's really just setting the 0th element of what becomes the global date array to the expansion of $1.

Having said I'll that, I'll show you a way to fake it using the declare built-in and indirect parameter expansion.

function assign () {
    ref=$1
    # Without the -g, we'd declare function-local parameters. The argument is a
    # string to evaluate as a variable assignment. If $ref=my_date1, then we do
    # 'my_date1[1]=$YY', 'my_date1[2]=$mm', etc.
    declare -g "$ref[1]=$YY"
    declare -g "$ref[2]=$mm"
    declare -g "$ref[3]=$dd"
}

function display () {
    ref=$1
    # If $ref=my_date1, then idx
    # iterates over my_date[1], my_date[2], my_date[3].
    # E.g. ${!idx} in the first iteration is ${my_date[1]}.
    arr=()
    for idx in $ref[{1,2,3}]; do
        arr+=( ${!idx} )
    done
    local IFS="/"
    echo "${arr[*]}"
}

export YY=2012 mm=09 dd=20
assign my_date1    # The *name* of the array; no need to predeclare

export YY=2012 mm=08 dd=20
assign my_date2    # The *name* of the array; no need to predeclare

# Again, just the *names* of the arrays
display my_date1
display my_date2

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