3

I would like to use a variable say:

i=1

as a value to refer to positional variables passed to script, e.g.:

x=101
y=201
z=301

foo(){
    echo "$1"
    echo "$2"
    echo "$3"
}

foo x y z

output:

101
201
301

Instead of refering to each parameter by index, how could I use i to increment through as a index variable?

To clarify:

foo() {
    local i=1
    echo "$i" #echo first paramter
    (( i+=1 ))
    echo "$i" #echo second parameter
    #etc.
}

what is the syntax for the echo "$i" part?

UPDATE after @Eric answer

~$ t=5
~$ foo() { i=1; echo "${!i}"; }
~$ foo t
t
~$

Update #2

So in short, the only way I can make my method work is by this:

foo() { #assuming 3 parameters

    i=0
    (( i+=1 ))
    var="${!i}"
    echo "${!var}"

    (( i+=1 ))
    var="${!i}"
    echo "${!var}"

    (( i+=1 ))
    var="${!i}"
    echo "${!var}"        
}
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3
  • 2
    Best to use arrays. But if you'd like the form ${!i} can be useful for you Commented Feb 29, 2016 at 19:38
  • Other way is to use shift command to operate arguments Commented Feb 29, 2016 at 19:40
  • Well, as the examples showed, you don't have to manually increment i yourself, but more or less. What is it that you're really trying to accomplish though? This feels like it could be a bit of an X-Y Problem Commented Feb 29, 2016 at 22:57

2 Answers 2

Reset to default
5

This is similar to the SO question here, which is similar to @costas comment. You can use $# to get the number of arguments and then indirect references like ${!i} to access a variable by name. Here's an example:

f() {
    for((i=1; i<=$#; i++)); do
        printf "%d %s\n" "$i" "${!i}"
    done
}

f a b c

which prints:

1 a
2 b
3 c

Seeing now that you want to pass in the names of variables as the positional arguments, you can an extra layer of indirect reference like so:

a=first
b=second
c=third

f() {
    for((i=1; i<=$#; i++)); do
        var="${!i}"
        printf "%d %s\n" "$i" "${!var}"
    done
}

f a b c

which prints

1 first
2 second
3 third

This lets us treat each argument as the name of a variable, which we store in var here. Then we access that variable indirectly in the printf.

You only get one layer of indirection at a time, nesting doesn't work. So trying to do it in one fell swoop as ${!${!i}} doesn't work because the first { starts the expansion, with the rest being treated as the PARAMETER value to expand. The first character being ! it will treat the rest as the name of the PARAMETER containing the name of the parameter we want, but ${!i} is not a valid parameter name, so we get a bad substitution. So we just do it in 2 steps to avoid that problem.

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4
  • see edit, this is does not seem to work for me Commented Feb 29, 2016 at 19:53
  • @Cybex See my update for the extra layer of indirection if you want to pass in the name of a variable to access Commented Feb 29, 2016 at 20:00
  • Thank you, this works, will mark as correct, however please explain the purpose of var="${!i}". When I tried combining the var and "print" variables from your edit, like this " echo "${!${!i}}" ", it fails, with a bad substitution. Could you elaborate on the var used and how it is used to display the containing data. Commented Feb 29, 2016 at 20:21
  • See update #2 please Commented Feb 29, 2016 at 20:46
2

Using shift, you can do the following:

num_args="$#"

for (( i = 1; i <= "$num_args"; i++ )); do
    echo "$1"
    shift
done

Here, the shift command shifts every next argument from the arguments passed to the script to be the value of $1.

For example:

$ ./myscript one two three four five six
one
two
three
four
five
six

whereby, in first run of the loop, $1 = "one", then in second run of loop, $1 = "two", so forth, until last argument.

Edit: If you want access to any of your script arguments without order, eg. using 2nd and 4th arguments at the same time, one method is using bash arrays as follows:

# Initialize the array 'myargs' with all script arguments
myargs=( "$@" )

for (( i = 0; i < "$#"; i++ )); do
    echo "${myargs[$i]}"
done

Here, you can access any of your arguments with any specified index i using the form "${myargs[$i]}".

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2
  • how would I retrieve variable #2 (or any other processed variable) when I shifted to current variable #4 ( or higher )? Commented Feb 29, 2016 at 19:57
  • Unfortunately, when you shift, your initial $1 values get lost unless you store them in another variable. Hence, the edit I made above, using an array to access any and all arguments freely any time in the script. Commented Feb 29, 2016 at 20:29

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