while passing some rgument with the "\*" char, my shell recognize it
as the "ls myfolder" command , and so print to the stdout the result of this command.

Any way to fix this problem? 

```
  1 #!/bin/bash
  2
  3 if [ $# -lt 3 ]; then
  4 echo "Some arguments are missing"
  5 fi
  6
  7 #passing "*" as argument 
  8 echo $2
  9 # the result is file1.txt file2.txt...and more..

```