Is it possible to first pass a parameter to the function and than change its value?
#!/bin/bash
name=old_name
echo $name #echoes "old_name"
alter () {
$1=new_name #throws error that says 'command not found'
}
alter name
echo $name #I would like to see "new_name" here
Yes, using a nameref:
alter () {
declare -n foo=$1
foo=new_name
}
See Bash FAQ 006 for more advice and warnings, as well as workarounds for versions of bash that predate nameref support (i.e., 4.2 or earlier).
You can't do it that way, unfortunately. Also, bash functions can't return values. The usual options are (1) set a global var in the function (yuck), or (2) echo the "return" value and use command substitution to call it. Something like this:
alter () {
echo "$1 but different"
}
name="Fred"
name=$(alter $name)
echo Name is now $name
returns: Name is now Fred but different
