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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            この記事は検証可能な参考文献や出典が全く示されていないか、不十分です。 出典を追加して記事の信頼性向上にご協力ください。 （ 2019年1月 ） 「 サソリ 」のその他の用法については「サソリ (曖昧さ回避)」をご覧ください。 「 スコーピオン 」と「 スコルピオン 」はこの項目へ転送されています。その他の用法については「スコーピオン」をご覧ください。 サソリ目 生息年代: 430–0 Ma PreЄ Є O S D C P T J K Pg N シルル紀前期-現世 ダイオウサソリ（ Pandinus Imperator ） 分類 界 : 動物界 Animalia 門 : 節足動物門 Arthropoda 亜門 : 鋏角亜門 Chelicerata 綱 : クモガタ綱 Arachnida 目 : サソリ目 Scorpiones 学名 Scorpiones C. L. Koch, 1837 英名 Scorpion 上科 Pseudochactoidea キョクトウサソリ上科 Buthoidea Chaeriloidea Chactoidea Iuroidea Scorpionoidea サソリ （ サソリ類 、 蠍 、 蝎 、 scorpion ）は、鋏角亜門・クモガタ綱の サソリ目 （ Scorpiones ）に属する節足動物の総称である。体の前端に鋏型の触肢、後端に毒針を有する捕食者である。 1700種以上を含め、最古の化石記録は4億3千万年前のシルル紀まで遡る。有毒生物として広く知られるが、人間の命にかかわる毒を持つものはごく一部の種類に過ぎない。 目次 1 形態 1.1 前体 1.2 後体 1.2.1 中体 1.2.2 終体 2 雌雄 3 生態 3.1 天敵 3.2 生殖 4 分布 5 毒 6 蛍光現象 7 分類
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            広島県道265号伴広島線

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            一般県道 広島県道265号伴広島線 路線延長 km 起点 広島県広島市安佐南区伴東7丁目 終点 広島県広島市中区榎町 接続する 主な道路 （記法） 広島県道38号広島豊平線 広島県道71号広島湯来線 山陽自動車道五日市IC 国道183号（ 国道261号重複） ■テンプレート（■ノート ■使い方） ■PJ道路 広島県道265号伴広島線 （ひろしまけんどう265ごう ともひろしません）は広島県広島市安佐南区から広島県広島市中区を結ぶ一般県道である。 目次 1 路線データ 2 通過する自治体 3 見どころ 4 接続道路 4.1 本線 4.2 旧道 5 別名 6 関連項目 路線データ 本線 起点：広島市安佐南区伴東7丁目・大原下橋南詰交差点（広島県道38号広島豊平線（本線・旧道）交点） 終点：広島市中区榎町・十日市交差点（国道183号（国道261号重用）交点） 総延長： 旧道 起点：広島市安佐南区伴東4丁目・新細坂交差点（広島県道38号広島豊平線交点） 終点：広島市安佐南区伴中央4丁目・寺組橋(南)交差点（広島県道265号伴広島線（本線）交点） 総延長： 通過する自治体 広島県 広島市（安佐南区、佐伯区、西区、中区） 見どころ 広島高速交通広島新交通1号線（アストラムライン） 広島広域公園 広島広域公園陸上競技場（広島ビッグアーチ） 己斐峠 己斐橋（太田川放水路） 天満橋（天満川） 接続道路 本線 広島県道38号広島豊平線（本線・旧道）（広島市安佐南区伴東7丁目・大原下橋南詰交差点〔起点〕） 広島県道71号広島湯来線（本線）（広島市安佐南区伴中央2丁目・伴交番入口交差点 - 広島市安佐南区伴中央4丁目・寺組橋(南)交差点間で重用） 広島県道71号広島湯来線（旧道）（広島県道265号伴広島線（旧道）重用）（広島市安佐南区伴中央4丁目・寺組橋(南)交差点） 広島県道71号広島湯来線（広島市安佐南区伴中央4丁目・寺組橋(南)交差点 - 広島市佐伯区五日市町石内・五月が丘(南)交差点間で重用）
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            Setup Asymptote in Texstudio

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            0 Good time of the day! I am trying to compile the following file: documentclass[12pt]{article} usepackage[inline]{asymptote} begin{document} Here is a venn diagram produced with Asymptote, drawn to width 4cm: defA{A} defB{V{B}} begin{figure} begin{center} begin{asy} size(4cm,0); pen colour1=red; pen colour2=green; pair z0=(0,0); pair z1=(-1,0); pair z2=(1,0); real r=1.5; path c1=circle(z1,r); path c2=circle(z2,r); fill(c1,colour1); fill(c2,colour2); picture intersection=new picture; fill(intersection,c1,colour1+colour2); clip(intersection,c2); add(intersection); draw(c1); draw(c2); //draw("$A$",box,z1)
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