Passing arguments by reference

Question

I want to ask if it is possible to pass arguments to a script function by reference:

i.e. to do something that would look like this in C++:

Answer 1

#!/bin/bash

append_string()
{
if [ -z "${!1}" ]; then
eval "${1}='$2'"
else
eval "${1}='${!1}''${!3}''$2'"
fi
}

PETS=''
SEP='|'
append_string "PETS" "cat" "SEP"
echo "$PETS"
append_string "PETS" "dog" "SEP"
echo "$PETS"
append_string "PETS" "hamster" "SEP"
echo "$PETS"

Output:

cat
cat|dog
cat|dog|hamster

Structure for calling that function is:

append_string  name_of_var_to_update  string_to_add  name_of_var_containing_sep_char

Name of variable is passed to fuction about PETS and SEP while string to append is passed the usual way as value. "${!1}" refers to contents of global PETS variable. In the beginning that variable is empty and contens is added each time we call the function. Separator character can be selected as needed. "eval" starting lines update PETS variable.

Answer 2

Ok, so this question has been waiting for a 'real' solution for some time now, and I am glad to say that we can now accomplish this without using eval at all.

The key to remember is to declare a reference in both the caller as the callee, at least in my example:

#!/bin/bash

# NOTE this does require a bash version >= 4.3

set -o errexit -o nounset -o posix -o pipefail

passedByRef() {

    local -n theRef

    if [ 0 -lt $# ]; then

        theRef=$1

        echo -e "${FUNCNAME}:\n\tthe value of my reference is:\n\t\t${theRef}"

        # now that we have a reference, we can assign things to it

        theRef="some other value"

        echo -e "${FUNCNAME}:\n\tvalue of my reference set to:\n\t\t${theRef}"

    else

        echo "Error: missing argument"

        exit 1
    fi
}

referenceTester() {

    local theVariable="I am a variable"

    # note the absence of quoting and escaping etc.

    local -n theReference=theVariable

    echo -e "${FUNCNAME}:\n\tthe value of my reference is:\n\t\t${theReference}"

    passedByRef theReference

    echo -e "${FUNCNAME}:\n\tthe value of my reference is now:\n\t\t${theReference},\n\tand the pointed to variable:\n\t\t${theVariable}"

}

# run it

referenceTester

Answer 3

I have found a way to do this but I am not sure how correct this is:

Newfun()
{
    local var1="$1"
    eval $var1=2
    # or can do eval $1=2 if no local var
}

var=1
echo  var is $var    # $var = 1
newfun 'var'         # pass the name of the variable…
echo now var is $var # $var = 2

So we pass the variable name as opposed to the value and then use eval ...

Answer 4

From the Bash man-page (Parameter Expansion):

    If the first  character of parameter is an exclamation  point (!), a
    level of variable indirection is  introduced. Bash uses the value of
    the variable  formed from the rest  of parameter as the  name of the
    variable; this variable  is then expanded and that value  is used in
    the rest  of the  substitution, rather than  the value  of parameter
    itself. This is known as indirect expansion.

Therefore a reference is the variable's name. Here is a swap function using variable indirection that does not require a temporary variable:

function swap()
{   # 
    # @param VARNAME1 VARNAME2
    #
    eval "$1=${!2} $2=${!1}"
}

$ a=1 b=2
$ swap a b
$ echo $a $b
2 1

Answer 5

Bash doesn't have anything like references built into it, so basically the only way you would be able to do what you want is to pass the function the name of the global variable you want it to modify. And even then you'll need an eval statement:

boo() {
    eval ${1}="new"
}

SOME_VAR="old"
echo $SOME_VAR # old
boo "SOME_VAR"
echo $SOME_VAR # new

I don't think you can use indirect references here because Bash automatically accesses the value of the variable whose name is stored in the indirect reference. It doesn't give you the chance to set it.

Answer 6

Use a helper function upvar:

# Assign variable one scope above the caller.
# Usage: local "$1" && upvar $1 value [value ...]
# Param: $1  Variable name to assign value to
# Param: $*  Value(s) to assign.  If multiple values, an array is
#            assigned, otherwise a single value is assigned.
# NOTE: For assigning multiple variables, use 'upvars'.  Do NOT
#       use multiple 'upvar' calls, since one 'upvar' call might
#       reassign a variable to be used by another 'upvar' call.
# See: http://fvue.nl/wiki/Bash:_Passing_variables_by_reference
upvar() {
    if unset -v "$1"; then           # Unset & validate varname
        if (( $# == 2 )); then
            eval $1=\"\$2\"          # Return single value
        else
            eval $1=\(\"\${@:2}\"\)  # Return array
         fi
    fi
}

And use it like this from within Newfun():

local "$1" && upvar $1 new

For returning multiple variables, use another helper function upvars. This allows passing multiple variables within one call, thus avoiding possible conflicts if one upvar call changes a variable used in another subsequent upvar call.

See: http://www.fvue.nl/wiki/Bash:_Passing_variables_by_reference for helper function upvars and more information.

The problem with:

eval $1=new

is that it's not safe if $1 happens to contain a command:

set -- 'ls /;true'
eval $1=new  # Oops

It would be better to use printf -v:

printf -v "$1" %s new

But printf -v cannot assign arrays.

Moreover, both eval and printf won't work if the variable happens to be declared local:

g() { local b; eval $1=bar; }  # WRONG
g b                            # Conflicts with `local b'
echo $b                        # b is empty unexpected

The conflict stays there even if local b is unset:

g() { local b; unset b; eval $1=bar; }  # WRONG
g b                                     # Still conflicts with `local b'
echo $b                                 # b is empty unexpected