[bash] Process all arguments except the first one (in a bash script)

Working in bash 4 or higher version:

#!/bin/bash
echo "$0";         #"bash"
bash --version;    #"GNU bash, version 5.0.3(1)-release (x86_64-pc-linux-gnu)"

In function:

echo $@;              #"p1" "p2" "p3" "p4" "p5"
echo ${@: 0};  #"bash" "p1" "p2" "p3" "p4" "p5"
echo ${@: 1};         #"p1" "p2" "p3" "p4" "p5"
echo ${@: 2};              #"p2" "p3" "p4" "p5"
echo ${@: 2:1};            #"p2"
echo ${@: 2:2};            #"p2" "p3"
echo ${@: -2};                       #"p4" "p5"
echo ${@: -2:1};                     #"p4"

Notice the space between ':' and '-', otherwise it means different:

${var:-word} If var is null or unset,
word is substituted for var. The value of var does not change.

${var:+word} If var is set,
word is substituted for var. The value of var does not change.

Which is described in:Unix / Linux - Shell Substitution