Picking specific array element











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I have just started reading about "$@" and "$*", I wanted to know if I can specifically point to an element in the "$@" array. Like without using any loop, I want to be able to pick element number 3 from "$@". Is there a way of doing this like "$1+@" or something like this? I already know about "${1}" but want to know specifically about "$@" and "$*". I tried searching for it but did not find anything related to this.






array







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  • 1




    array=(apple banana orange); printf "%sn" "${array[1]}"
    – jasonwryan
    Jul 12 '17 at 0:40












  • @jasonwryan Thanks, it worked, doesn't any syntax something specifically like "$1+@" or "$@+1" exist? I think I saw it somewhere, but am not sure because it has been a lot of time.
    – GypsyCosmonaut
    Jul 12 '17 at 1:02

















up vote
1
down vote

favorite












I have just started reading about "$@" and "$*", I wanted to know if I can specifically point to an element in the "$@" array. Like without using any loop, I want to be able to pick element number 3 from "$@". Is there a way of doing this like "$1+@" or something like this? I already know about "${1}" but want to know specifically about "$@" and "$*". I tried searching for it but did not find anything related to this.






array







share|improve this question




















  • 1




    array=(apple banana orange); printf "%sn" "${array[1]}"
    – jasonwryan
    Jul 12 '17 at 0:40












  • @jasonwryan Thanks, it worked, doesn't any syntax something specifically like "$1+@" or "$@+1" exist? I think I saw it somewhere, but am not sure because it has been a lot of time.
    – GypsyCosmonaut
    Jul 12 '17 at 1:02















up vote
1
down vote

favorite









up vote
1
down vote

favorite











I have just started reading about "$@" and "$*", I wanted to know if I can specifically point to an element in the "$@" array. Like without using any loop, I want to be able to pick element number 3 from "$@". Is there a way of doing this like "$1+@" or something like this? I already know about "${1}" but want to know specifically about "$@" and "$*". I tried searching for it but did not find anything related to this.






array







share|improve this question















I have just started reading about "$@" and "$*", I wanted to know if I can specifically point to an element in the "$@" array. Like without using any loop, I want to be able to pick element number 3 from "$@". Is there a way of doing this like "$1+@" or something like this? I already know about "${1}" but want to know specifically about "$@" and "$*". I tried searching for it but did not find anything related to this.






array




array






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited yesterday









Rui F Ribeiro

38.6k1479128




38.6k1479128










asked Jul 12 '17 at 0:36









GypsyCosmonaut

758928




758928








  • 1




    array=(apple banana orange); printf "%sn" "${array[1]}"
    – jasonwryan
    Jul 12 '17 at 0:40












  • @jasonwryan Thanks, it worked, doesn't any syntax something specifically like "$1+@" or "$@+1" exist? I think I saw it somewhere, but am not sure because it has been a lot of time.
    – GypsyCosmonaut
    Jul 12 '17 at 1:02
















  • 1




    array=(apple banana orange); printf "%sn" "${array[1]}"
    – jasonwryan
    Jul 12 '17 at 0:40












  • @jasonwryan Thanks, it worked, doesn't any syntax something specifically like "$1+@" or "$@+1" exist? I think I saw it somewhere, but am not sure because it has been a lot of time.
    – GypsyCosmonaut
    Jul 12 '17 at 1:02










1




1




array=(apple banana orange); printf "%sn" "${array[1]}"
– jasonwryan
Jul 12 '17 at 0:40






array=(apple banana orange); printf "%sn" "${array[1]}"
– jasonwryan
Jul 12 '17 at 0:40














@jasonwryan Thanks, it worked, doesn't any syntax something specifically like "$1+@" or "$@+1" exist? I think I saw it somewhere, but am not sure because it has been a lot of time.
– GypsyCosmonaut
Jul 12 '17 at 1:02






@jasonwryan Thanks, it worked, doesn't any syntax something specifically like "$1+@" or "$@+1" exist? I think I saw it somewhere, but am not sure because it has been a lot of time.
– GypsyCosmonaut
Jul 12 '17 at 1:02












3 Answers
3






active

oldest

votes

















up vote
1
down vote



accepted










It is said that the positional parameters are not an array.

And the way that exists to set them is via set. No other array needs that.



$ set -- one two t33 f44

$ printf '%sn' "$@"

one

two

t33

f44


But at least in bash (and ksh and zsh), they could be selected just as easy:



$ set -- one two t33 f44

$ echo "${@:2:1}"

two

$ echo "${@:2:2}"

two t33





share|improve this answer





















  • If echo echo ${@:2:1} gives the output two. So does this mean that one two t33 f44 are set at the indexes 1 2 3 4 respectively instead of usual format 0 1 2 3 considered in arrays?
    – GypsyCosmonaut
    Jul 12 '17 at 1:24












  • And also, why does echo ${@:0:1} gives the output similar to echo $0 i.e the current shell name?
    – GypsyCosmonaut
    Jul 12 '17 at 1:30






  • 1




    Yes!. That's the usual numbering of positional parameters as $0 is (most of the time) the name of the running script/program. Just do an echo $0 to see it. @GypsyCosmonaut
    – Arrow
    Jul 12 '17 at 1:31






  • 1




    Well, as the zero index is the name of the executing script/shell, it must be what should be printed by asking for it with echo ${@:0:1}. Doesn't it? @GypsyCosmonaut
    – Arrow
    Jul 12 '17 at 1:33






  • 1




    ${parameter:offset:length} is substring expansion in general, but for @ it is treated differently: ${@:offset:length}you get length positional parameters beginning at offset.
    – NickD
    Jul 12 '17 at 1:34


















up vote
1
down vote













$@ is not an array: it's just a list of the arguments. In bash, you can create an array, initialize it with the values from $@ and then use indexing:



declare -a foo=($@)



echo ${foo[2]}


The array indices start from 0, so the above prints the third argument to the script.






share|improve this answer




























    up vote
    0
    down vote













    Essentially, you can't. The thing is that $* and $@ are not arrays; they are simple variables. Thus, it isn't possible to index them.



    Their values are just strings, defined in slightly different ways. $1, $2, etc. give you access to the individual components.






    share|improve this answer

















    • 1




      I believe that you should take a look to my answer. There is an easy way of using indexes to read elements of the positional parameters (at least on some shells)
      – Arrow
      Jul 12 '17 at 1:17










    • My answer does not conflict with that. I was correcting the assumption in the question that the parameters are arrays, that's all.
      – Bob Eager
      Jul 12 '17 at 6:38











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    3 Answers
    3






    active

    oldest

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    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    It is said that the positional parameters are not an array.

    And the way that exists to set them is via set. No other array needs that.



    $ set -- one two t33 f44
    
$ printf '%sn' "$@"
    
one
    
two
    
t33
    
f44


    But at least in bash (and ksh and zsh), they could be selected just as easy:



    $ set -- one two t33 f44
    
$ echo "${@:2:1}"
    
two
    
$ echo "${@:2:2}"
    
two t33





    share|improve this answer





















    • If echo echo ${@:2:1} gives the output two. So does this mean that one two t33 f44 are set at the indexes 1 2 3 4 respectively instead of usual format 0 1 2 3 considered in arrays?
      – GypsyCosmonaut
      Jul 12 '17 at 1:24












    • And also, why does echo ${@:0:1} gives the output similar to echo $0 i.e the current shell name?
      – GypsyCosmonaut
      Jul 12 '17 at 1:30






    • 1




      Yes!. That's the usual numbering of positional parameters as $0 is (most of the time) the name of the running script/program. Just do an echo $0 to see it. @GypsyCosmonaut
      – Arrow
      Jul 12 '17 at 1:31






    • 1




      Well, as the zero index is the name of the executing script/shell, it must be what should be printed by asking for it with echo ${@:0:1}. Doesn't it? @GypsyCosmonaut
      – Arrow
      Jul 12 '17 at 1:33






    • 1




      ${parameter:offset:length} is substring expansion in general, but for @ it is treated differently: ${@:offset:length}you get length positional parameters beginning at offset.
      – NickD
      Jul 12 '17 at 1:34















    up vote
    1
    down vote



    accepted










    It is said that the positional parameters are not an array.

    And the way that exists to set them is via set. No other array needs that.



    $ set -- one two t33 f44
    
$ printf '%sn' "$@"
    
one
    
two
    
t33
    
f44


    But at least in bash (and ksh and zsh), they could be selected just as easy:



    $ set -- one two t33 f44
    
$ echo "${@:2:1}"
    
two
    
$ echo "${@:2:2}"
    
two t33





    share|improve this answer





















    • If echo echo ${@:2:1} gives the output two. So does this mean that one two t33 f44 are set at the indexes 1 2 3 4 respectively instead of usual format 0 1 2 3 considered in arrays?
      – GypsyCosmonaut
      Jul 12 '17 at 1:24












    • And also, why does echo ${@:0:1} gives the output similar to echo $0 i.e the current shell name?
      – GypsyCosmonaut
      Jul 12 '17 at 1:30






    • 1




      Yes!. That's the usual numbering of positional parameters as $0 is (most of the time) the name of the running script/program. Just do an echo $0 to see it. @GypsyCosmonaut
      – Arrow
      Jul 12 '17 at 1:31






    • 1




      Well, as the zero index is the name of the executing script/shell, it must be what should be printed by asking for it with echo ${@:0:1}. Doesn't it? @GypsyCosmonaut
      – Arrow
      Jul 12 '17 at 1:33






    • 1




      ${parameter:offset:length} is substring expansion in general, but for @ it is treated differently: ${@:offset:length}you get length positional parameters beginning at offset.
      – NickD
      Jul 12 '17 at 1:34













    up vote
    1
    down vote



    accepted







    up vote
    1
    down vote



    accepted






    It is said that the positional parameters are not an array.

    And the way that exists to set them is via set. No other array needs that.



    $ set -- one two t33 f44
    
$ printf '%sn' "$@"
    
one
    
two
    
t33
    
f44


    But at least in bash (and ksh and zsh), they could be selected just as easy:



    $ set -- one two t33 f44
    
$ echo "${@:2:1}"
    
two
    
$ echo "${@:2:2}"
    
two t33





    share|improve this answer












    It is said that the positional parameters are not an array.

    And the way that exists to set them is via set. No other array needs that.



    $ set -- one two t33 f44
    
$ printf '%sn' "$@"
    
one
    
two
    
t33
    
f44


    But at least in bash (and ksh and zsh), they could be selected just as easy:



    $ set -- one two t33 f44
    
$ echo "${@:2:1}"
    
two
    
$ echo "${@:2:2}"
    
two t33






    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Jul 12 '17 at 1:14









    Arrow

    2,460218




    2,460218












    • If echo echo ${@:2:1} gives the output two. So does this mean that one two t33 f44 are set at the indexes 1 2 3 4 respectively instead of usual format 0 1 2 3 considered in arrays?
      – GypsyCosmonaut
      Jul 12 '17 at 1:24












    • And also, why does echo ${@:0:1} gives the output similar to echo $0 i.e the current shell name?
      – GypsyCosmonaut
      Jul 12 '17 at 1:30






    • 1




      Yes!. That's the usual numbering of positional parameters as $0 is (most of the time) the name of the running script/program. Just do an echo $0 to see it. @GypsyCosmonaut
      – Arrow
      Jul 12 '17 at 1:31






    • 1




      Well, as the zero index is the name of the executing script/shell, it must be what should be printed by asking for it with echo ${@:0:1}. Doesn't it? @GypsyCosmonaut
      – Arrow
      Jul 12 '17 at 1:33






    • 1




      ${parameter:offset:length} is substring expansion in general, but for @ it is treated differently: ${@:offset:length}you get length positional parameters beginning at offset.
      – NickD
      Jul 12 '17 at 1:34


















    • If echo echo ${@:2:1} gives the output two. So does this mean that one two t33 f44 are set at the indexes 1 2 3 4 respectively instead of usual format 0 1 2 3 considered in arrays?
      – GypsyCosmonaut
      Jul 12 '17 at 1:24












    • And also, why does echo ${@:0:1} gives the output similar to echo $0 i.e the current shell name?
      – GypsyCosmonaut
      Jul 12 '17 at 1:30






    • 1




      Yes!. That's the usual numbering of positional parameters as $0 is (most of the time) the name of the running script/program. Just do an echo $0 to see it. @GypsyCosmonaut
      – Arrow
      Jul 12 '17 at 1:31






    • 1




      Well, as the zero index is the name of the executing script/shell, it must be what should be printed by asking for it with echo ${@:0:1}. Doesn't it? @GypsyCosmonaut
      – Arrow
      Jul 12 '17 at 1:33






    • 1




      ${parameter:offset:length} is substring expansion in general, but for @ it is treated differently: ${@:offset:length}you get length positional parameters beginning at offset.
      – NickD
      Jul 12 '17 at 1:34
















    If echo echo ${@:2:1} gives the output two. So does this mean that one two t33 f44 are set at the indexes 1 2 3 4 respectively instead of usual format 0 1 2 3 considered in arrays?
    – GypsyCosmonaut
    Jul 12 '17 at 1:24






    If echo echo ${@:2:1} gives the output two. So does this mean that one two t33 f44 are set at the indexes 1 2 3 4 respectively instead of usual format 0 1 2 3 considered in arrays?
    – GypsyCosmonaut
    Jul 12 '17 at 1:24














    And also, why does echo ${@:0:1} gives the output similar to echo $0 i.e the current shell name?
    – GypsyCosmonaut
    Jul 12 '17 at 1:30




    And also, why does echo ${@:0:1} gives the output similar to echo $0 i.e the current shell name?
    – GypsyCosmonaut
    Jul 12 '17 at 1:30




    1




    1




    Yes!. That's the usual numbering of positional parameters as $0 is (most of the time) the name of the running script/program. Just do an echo $0 to see it. @GypsyCosmonaut
    – Arrow
    Jul 12 '17 at 1:31




    Yes!. That's the usual numbering of positional parameters as $0 is (most of the time) the name of the running script/program. Just do an echo $0 to see it. @GypsyCosmonaut
    – Arrow
    Jul 12 '17 at 1:31




    1




    1




    Well, as the zero index is the name of the executing script/shell, it must be what should be printed by asking for it with echo ${@:0:1}. Doesn't it? @GypsyCosmonaut
    – Arrow
    Jul 12 '17 at 1:33




    Well, as the zero index is the name of the executing script/shell, it must be what should be printed by asking for it with echo ${@:0:1}. Doesn't it? @GypsyCosmonaut
    – Arrow
    Jul 12 '17 at 1:33




    1




    1




    ${parameter:offset:length} is substring expansion in general, but for @ it is treated differently: ${@:offset:length}you get length positional parameters beginning at offset.
    – NickD
    Jul 12 '17 at 1:34




    ${parameter:offset:length} is substring expansion in general, but for @ it is treated differently: ${@:offset:length}you get length positional parameters beginning at offset.
    – NickD
    Jul 12 '17 at 1:34












    up vote
    1
    down vote













    $@ is not an array: it's just a list of the arguments. In bash, you can create an array, initialize it with the values from $@ and then use indexing:



    declare -a foo=($@)
    

    
echo ${foo[2]}


    The array indices start from 0, so the above prints the third argument to the script.






    share|improve this answer

























      up vote
      1
      down vote













      $@ is not an array: it's just a list of the arguments. In bash, you can create an array, initialize it with the values from $@ and then use indexing:



      declare -a foo=($@)
      

      
echo ${foo[2]}


      The array indices start from 0, so the above prints the third argument to the script.






      share|improve this answer























        up vote
        1
        down vote










        up vote
        1
        down vote









        $@ is not an array: it's just a list of the arguments. In bash, you can create an array, initialize it with the values from $@ and then use indexing:



        declare -a foo=($@)
        

        
echo ${foo[2]}


        The array indices start from 0, so the above prints the third argument to the script.






        share|improve this answer












        $@ is not an array: it's just a list of the arguments. In bash, you can create an array, initialize it with the values from $@ and then use indexing:



        declare -a foo=($@)
        

        
echo ${foo[2]}


        The array indices start from 0, so the above prints the third argument to the script.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Jul 12 '17 at 1:02









        NickD

        1,6551313




        1,6551313






















            up vote
            0
            down vote













            Essentially, you can't. The thing is that $* and $@ are not arrays; they are simple variables. Thus, it isn't possible to index them.



            Their values are just strings, defined in slightly different ways. $1, $2, etc. give you access to the individual components.






            share|improve this answer

















            • 1




              I believe that you should take a look to my answer. There is an easy way of using indexes to read elements of the positional parameters (at least on some shells)
              – Arrow
              Jul 12 '17 at 1:17










            • My answer does not conflict with that. I was correcting the assumption in the question that the parameters are arrays, that's all.
              – Bob Eager
              Jul 12 '17 at 6:38















            up vote
            0
            down vote













            Essentially, you can't. The thing is that $* and $@ are not arrays; they are simple variables. Thus, it isn't possible to index them.



            Their values are just strings, defined in slightly different ways. $1, $2, etc. give you access to the individual components.






            share|improve this answer

















            • 1




              I believe that you should take a look to my answer. There is an easy way of using indexes to read elements of the positional parameters (at least on some shells)
              – Arrow
              Jul 12 '17 at 1:17










            • My answer does not conflict with that. I was correcting the assumption in the question that the parameters are arrays, that's all.
              – Bob Eager
              Jul 12 '17 at 6:38













            up vote
            0
            down vote










            up vote
            0
            down vote









            Essentially, you can't. The thing is that $* and $@ are not arrays; they are simple variables. Thus, it isn't possible to index them.



            Their values are just strings, defined in slightly different ways. $1, $2, etc. give you access to the individual components.






            share|improve this answer












            Essentially, you can't. The thing is that $* and $@ are not arrays; they are simple variables. Thus, it isn't possible to index them.



            Their values are just strings, defined in slightly different ways. $1, $2, etc. give you access to the individual components.







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Jul 12 '17 at 1:05









            Bob Eager

            1,8861421




            1,8861421








            • 1




              I believe that you should take a look to my answer. There is an easy way of using indexes to read elements of the positional parameters (at least on some shells)
              – Arrow
              Jul 12 '17 at 1:17










            • My answer does not conflict with that. I was correcting the assumption in the question that the parameters are arrays, that's all.
              – Bob Eager
              Jul 12 '17 at 6:38














            • 1




              I believe that you should take a look to my answer. There is an easy way of using indexes to read elements of the positional parameters (at least on some shells)
              – Arrow
              Jul 12 '17 at 1:17










            • My answer does not conflict with that. I was correcting the assumption in the question that the parameters are arrays, that's all.
              – Bob Eager
              Jul 12 '17 at 6:38








            1




            1




            I believe that you should take a look to my answer. There is an easy way of using indexes to read elements of the positional parameters (at least on some shells)
            – Arrow
            Jul 12 '17 at 1:17




            I believe that you should take a look to my answer. There is an easy way of using indexes to read elements of the positional parameters (at least on some shells)
            – Arrow
            Jul 12 '17 at 1:17












            My answer does not conflict with that. I was correcting the assumption in the question that the parameters are arrays, that's all.
            – Bob Eager
            Jul 12 '17 at 6:38




            My answer does not conflict with that. I was correcting the assumption in the question that the parameters are arrays, that's all.
            – Bob Eager
            Jul 12 '17 at 6:38


















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