Access arguments to Bash script inside a function [duplicate]

Question

Inside a function, $1 ... $n are the parameters passed to that function. Outside a function $1 ... $n are the parameters passed to the script.

Can I somehow access the parameters passed to the script inside a function?

Answer 1

Usually you just pass them as parameters to the function at call time.

The (uglier) alternative is to put them in global variables.

Answer 2

(I know this is an old post, but none of the answers actually answered the question.)

Use the BASH_ARGV array. It contains the arguments passed to the invoking script in reverse order (i.e., it's a stack with the top at index 0). You may have to turn on extended debugging in the shebang (e.g., #!/bin/bash -O extdebug) or with shopt (e.g., shopt -s extdebug), but it works for me in bash 4.2_p37 without it turned on.

From man bash:

An array variable containing all of the parameters in the current bash execution call stack. The final parameter of the last subroutine call is at the top of the stack; the first parameter of the initial call is at the bottom. When a subroutine is executed, the parameters supplied are pushed onto BASH_ARGV. The shell sets BASH_ARGV only when in extended debugging mode….

Here's a function I use to print all arguments in order on a single line:

# Print the arguments of the calling script, in order. function get_script_args {     # Get the number of arguments passed to this script.     # (The BASH_ARGV array does not include $0.)     local n=${#BASH_ARGV[@]}      if (( $n > 0 ))     then         # Get the last index of the args in BASH_ARGV.         local n_index=$(( $n - 1 ))          # Loop through the indexes from largest to smallest.         for i in $(seq ${n_index} -1 0)         do             # Print a space if necessary.             if (( $i < $n_index ))             then                 echo -n ' '             fi              # Print the actual argument.             echo -n "${BASH_ARGV[$i]}"         done          # Print a newline.         echo     fi }