Declaring global variable inside a function

Question

What I want to do is following. Inside a function, I need to assign a value to a variable, whose name is taken from another variable. In other words:

func() {     #     # Here happens something that ultimately makes $arg="var_name"   #    declare -g ${arg}=5 }  func  echo ${var_name}; # Prints out "5" 

The code snippet above works great in bash 4.2. However, in bash before 4.2, declare doesn't have the -g option. Everything I found at google says that to define the global variable inside a function, I should just use the var=value syntax, but unfortunately var itself depends on another variable. ${arg}=5 doesn't work, either. (It says -bash: var_name=5: command not found.

For the curious, the reason for all this is that this function actually creates global variables from the script parameters, i.e. running script --arg1=val automatically creates variable named arg1 with value val. Saves tons of a boilerplate code.

Answer 1

declare inside a function doesn't work as expected. I needed read-only global variables declared in a function. I tried this inside a function but it didn't work:

declare -r MY_VAR=1 

But this didn't work. Using the readonly command did:

func() {     readonly MY_VAR=1 } func echo $MY_VAR MY_VAR=2 

This will print 1 and give the error "MY_VAR: readonly variable" for the second assignment.