Question

I am changing a bash script which has a structure as such:

#somewhere in the code
sim_counts=#... some value
function_name()
{
    set $sim_counts
    for hostname in $linux_hostnames; do
        if [ $1 -eq 0 ]; then # if sim_counts equal 0
            shift  # jump forward in sim_counts
            continue 
        fi
    # ... more code
    shift
    done
}

Then it is called in the script:

function_name

---

I want to introduce a parameter to this function:

#somewhere in the code
sim_counts=#... some value
function_name()
{
    ip=$1
    set $sim_counts
    for hostname in $linux_hostnames; do
        if [ $1 -eq 0 ]; then # if sim_counts equal 0
            shift  # jump forward in sim_counts
            continue 
        fi
    # ... more code
    shift
    done
}

And call the function in following way:

function_name 10.255.192.123

What should I do to avoid $1 conflict of function parameter and the other value from set command ?

Answer 1

If I am correctly reading the set builtin page in the Bash Reference Manual, I believe the code as you have written it will just work. Quoting from that page:

The remaining N arguments are positional parameters and are assigned, in order, to $1, $2, … $N. The special parameter # is set to N.

In essence, any pre-existing values for the positional variables will be blown away. The first sentence on that manual page is also interesting:

This builtin is so complicated that it deserves its own section.

In short, I think your code should simply work as expected. You’ve saved the initial value of $1 (from the function call) into a temporary variable; as long as you refer to $ip for that specific value, you should be good. In my own test script, it seems that $1 gets blown away as I expect it should.