How to access command line arguments of the caller inside a function

Question

I m attempting to write a function in bash that will access the scripts command line arguments  but they are replaced with the positional arguments to the function  Is there any way for the function to access the command line arguments if they aren t passed in explicitly     Demo function function stuff     echo  0         Echo s the name of the script  but no command line arguments stuff    Echo s everything I want  but trying to avoid stuff

User · Answer

Save the script arguments SCRIPT NAME  0 ARG 1  1 ARGS ALL     function stuff       use script args via the variables you saved     or the function args via     echo  0           Call the function with arguments stuff 1 2 3 4

User · Answer

My solution   Create a function script that is called earlier than all other functions without passing any arguments to it  like this         bin bash      function init        ORIGOPT   -    -         Afer that  you can call init and use the ORIGOPT var as needed as a plus  I always assign a new var and copy the contents of ORIGOPT in my new functions  that way you can keep yourself assured nobody is going to touch it or change it   I added spaces and dashes to make it easier to parse it with  sed -E  also bash will not pass it as reference and make ORIGOPT grow as functions are called with more arguments

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usr bin env bash  echo name of script is  0 echo first argument is  1 echo second argument is  2 echo seventeenth argument is  17 echo number of arguments is      Edit  please see my comment on question

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If you want to have your arguments C style  array of arguments   number of arguments  you can use    and         gives you the number of arguments     gives you all arguments  You can turn this into an array by args          So for example   args        echo    arguments passed echo   args 0     args 1     args 2     Note that here   args 0   actually is the 1st argument and not the name of your script

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One can do it like this as well     bin bash   script name function test sh function argument    for i in    do     echo  i done    argument      Now call your script like    function test sh argument1 argument2

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You can use the shift keyword  operator   to iterate through them  Example      bin bash function print         while      -gt 0       do         echo  1          shift 1      done   print

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My reading of the Bash Reference Manual says this stuff is captured in BASH ARGV  although it talks about  quot the stack quot  a lot     bin bash  function argv       for a in   BASH ARGV       do       echo -n  quot  a  quot      done     echo    function f       echo f  1  2  3     echo -n f   argv    function g       echo g  1  2  3     echo -n g  argv     f    f boo bar baz g goo gar gaz  Save in f sh     f sh arg0 arg1 arg2 f boo bar baz farg2 arg1 arg0  g goo gar gaz garg2 arg1 arg0  f farg2 arg1 arg0

User · Answer

Ravi s comment is essentially the answer  Functions take their own arguments  If you want them to be the same as the command-line arguments  you must pass them in  Otherwise  you re clearly calling a function without arguments   That said  you could if you like store the command-line arguments in a global array to use within other functions   my function         echo  stored arguments       for arg in    commandline args       do         echo       arg      done    commandline args         my function   You have to access the command-line arguments through the commandline args variable  not      1   2  etc   but they re available  I m unaware of any way to assign directly to the argument array  but if someone knows one  please enlighten me   Also  note the way I ve used and quoted    - this is how you ensure special characters   whitespace  don t get mucked up

[bash] How to access command line arguments of the caller inside a function?

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