Access file-level argv from within a function

Question

I have a script, like follows:

echo 'Hello world' # etc

function fun
   if test 1 -gt 0
      echo "$argv[1]"
   end
end

# lots of code in between

fun

I call this file with ./file foo --bar. I want fun to print --bar. Instead, it prints nothing, because the argv containing --bar (ie, the file's argv) is being masked by the function's argv, which isn't receiving any input.

I could, theoretically, pass argv to the function, as in fun $argv. However, this is very obnoxious to have to include with every single function call (I make extensive references to argv in this file). Is there a way to access the file-scope argv variable from within functions in the file?

Answer 1

The outer argv is shadowed by the inner argv - a global variable shadowed by a local variable. There is no way to change this.

There are numerous ways around it, including what I would recommend: Use set -g fileargv $argv once before you execute the functions and then use $fileargv in your functions.