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difference between echo ($x) and a=“($x)”; echo $a



2019 Community Moderator ElectionWhen is double-quoting necessary?What's meaning the of a pair of parentheses after a string literal?For loop brackets - C like syntaxUnderstanding Bash's Read-a-File Command SubstitutionWhy is this simple bash script throwing an error on if/then/else?echo variable with content from command substitutionC Shell Script syntax error “unexpected end of of file”Bash: Syntax error - Unexpected near token 'else'Centos gawk command line usageblocking/non-blocking pipes/redirects inside command substitutionsyntax error near unexpected token `do










0















When I execute the following in terminal:



x=1
a=($x)

echo $a #output: 1
echo ($x) #output: -bash: syntax error near unexpected token `$x'


Why are above outputs different?
Also why are the following 2 outputs different:



$(echo foo) #output: -bash: foo: command not found
(echo foo) #output: foo


P.S.: I am new to shell scripting and trying to understand command substitution: $(command)










share|improve this question







New contributor




Sunny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
























    0















    When I execute the following in terminal:



    x=1
    a=($x)

    echo $a #output: 1
    echo ($x) #output: -bash: syntax error near unexpected token `$x'


    Why are above outputs different?
    Also why are the following 2 outputs different:



    $(echo foo) #output: -bash: foo: command not found
    (echo foo) #output: foo


    P.S.: I am new to shell scripting and trying to understand command substitution: $(command)










    share|improve this question







    New contributor




    Sunny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






















      0












      0








      0








      When I execute the following in terminal:



      x=1
      a=($x)

      echo $a #output: 1
      echo ($x) #output: -bash: syntax error near unexpected token `$x'


      Why are above outputs different?
      Also why are the following 2 outputs different:



      $(echo foo) #output: -bash: foo: command not found
      (echo foo) #output: foo


      P.S.: I am new to shell scripting and trying to understand command substitution: $(command)










      share|improve this question







      New contributor




      Sunny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.












      When I execute the following in terminal:



      x=1
      a=($x)

      echo $a #output: 1
      echo ($x) #output: -bash: syntax error near unexpected token `$x'


      Why are above outputs different?
      Also why are the following 2 outputs different:



      $(echo foo) #output: -bash: foo: command not found
      (echo foo) #output: foo


      P.S.: I am new to shell scripting and trying to understand command substitution: $(command)







      bash shell scripting command-substitution






      share|improve this question







      New contributor




      Sunny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|improve this question







      New contributor




      Sunny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|improve this question




      share|improve this question






      New contributor




      Sunny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked 20 hours ago









      SunnySunny

      31




      31




      New contributor




      Sunny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      Sunny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      Sunny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.




















          1 Answer
          1






          active

          oldest

          votes


















          4














          You are mixing the three different usages of parentheses in bash: array definition, command substitution, and command grouping. Command substitution and command grouping is standard syntax and all POSIX sh-like shells supports it. The array definition syntax is supported in shells that additionally have arrays.



          Array Definition




          Arrays are assigned to using compound assignments of the form



          name=(value1 value2 … )




          Your a=($x) creates an array with a single element, whose value is the value specified by $x, i.e. 1. When you echo an array by its name, in your case echo $a, it only echoes the first element of the array. So that's what you see, a 1 is printed to stdout.



          If you have multiple elements in an array you can use "$arrayname[*]" or "$arrayname[@]" to access them. The first combines all array elements into a single argument while the second put each element as an individual argument.



          This syntax is specific to bash and to other shells that allows for arrays in a similar way.



          Command substitution (POSIX)




          Command substitution allows the output of a command to replace the
          command itself. Command substitution occurs when a command is enclosed
          as follows:



          $(command)




          Your command $(echo foo) performs shell substitution and is parsed to foo. So it is the same as you have typed foo. Since the command foo does not exist, bash complains about it.



          Command grouping (POSIX)




          Bash provides two ways to group a list of commands to be executed as a
          unit. When commands are grouped, redirections may be applied to the
          entire command list. For example, the output of all the commands in
          the list may be redirected to a single stream.



          ( list )



          Placing a list of commands between parentheses causes a
          subshell environment to be created (see Command Execution
          Environment), and each of the commands in list to be executed in that
          subshell. Since the list is executed in a subshell, variable
          assignments do not remain in effect after the subshell completes.




          Your command (echo foo) executes echo foo in a subshell, so foo is echoed. This has nothing to do with the command substitution. As the manual says, variable assignments in subshell do not remain in effect after the subshell completes. This can come handy if you want to write simple one-liners. For example, instead of



          for l in 1..10; do mycommand $l; done; unset l


          You can write,



          ( for l in 1..10; do mycommand $l; done )


          Another helpful usage of a subshell is like using



          ( cd folder; ./mycommand )


          instead of



          cd folder; ./mycommand; cd -


          Your command echo ($x) falls into none of these three categories, and bash reports a syntax error.



          Quote your variables properly



          It is also worth mentioning that you should quote your variable expansions whenever applicable (for example, when a string should contain the value of a variable). See When is double-quoting necessary? for details.



          For example, the following would not invoke the particular constructs described above:



          x=1
          a="($x)"

          echo "$a"
          echo "($x)"


          Here, both calls to echo would output (1).



          Links to bash manual pages



          https://www.gnu.org/software/bash/manual/html_node/Command-Grouping.html



          https://www.gnu.org/software/bash/manual/html_node/Command-Substitution.html



          https://www.gnu.org/software/bash/manual/html_node/Arrays.html






          share|improve this answer




















          • 2





            Thank you. I've revised a bit but feel free to make edits on it. I am not that good at guiding new users.

            – Weijun Zhou
            19 hours ago











          • thanks for the detailed explanation Weijun

            – Sunny
            5 hours ago










          Your Answer








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          1 Answer
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          1 Answer
          1






          active

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          active

          oldest

          votes






          active

          oldest

          votes









          4














          You are mixing the three different usages of parentheses in bash: array definition, command substitution, and command grouping. Command substitution and command grouping is standard syntax and all POSIX sh-like shells supports it. The array definition syntax is supported in shells that additionally have arrays.



          Array Definition




          Arrays are assigned to using compound assignments of the form



          name=(value1 value2 … )




          Your a=($x) creates an array with a single element, whose value is the value specified by $x, i.e. 1. When you echo an array by its name, in your case echo $a, it only echoes the first element of the array. So that's what you see, a 1 is printed to stdout.



          If you have multiple elements in an array you can use "$arrayname[*]" or "$arrayname[@]" to access them. The first combines all array elements into a single argument while the second put each element as an individual argument.



          This syntax is specific to bash and to other shells that allows for arrays in a similar way.



          Command substitution (POSIX)




          Command substitution allows the output of a command to replace the
          command itself. Command substitution occurs when a command is enclosed
          as follows:



          $(command)




          Your command $(echo foo) performs shell substitution and is parsed to foo. So it is the same as you have typed foo. Since the command foo does not exist, bash complains about it.



          Command grouping (POSIX)




          Bash provides two ways to group a list of commands to be executed as a
          unit. When commands are grouped, redirections may be applied to the
          entire command list. For example, the output of all the commands in
          the list may be redirected to a single stream.



          ( list )



          Placing a list of commands between parentheses causes a
          subshell environment to be created (see Command Execution
          Environment), and each of the commands in list to be executed in that
          subshell. Since the list is executed in a subshell, variable
          assignments do not remain in effect after the subshell completes.




          Your command (echo foo) executes echo foo in a subshell, so foo is echoed. This has nothing to do with the command substitution. As the manual says, variable assignments in subshell do not remain in effect after the subshell completes. This can come handy if you want to write simple one-liners. For example, instead of



          for l in 1..10; do mycommand $l; done; unset l


          You can write,



          ( for l in 1..10; do mycommand $l; done )


          Another helpful usage of a subshell is like using



          ( cd folder; ./mycommand )


          instead of



          cd folder; ./mycommand; cd -


          Your command echo ($x) falls into none of these three categories, and bash reports a syntax error.



          Quote your variables properly



          It is also worth mentioning that you should quote your variable expansions whenever applicable (for example, when a string should contain the value of a variable). See When is double-quoting necessary? for details.



          For example, the following would not invoke the particular constructs described above:



          x=1
          a="($x)"

          echo "$a"
          echo "($x)"


          Here, both calls to echo would output (1).



          Links to bash manual pages



          https://www.gnu.org/software/bash/manual/html_node/Command-Grouping.html



          https://www.gnu.org/software/bash/manual/html_node/Command-Substitution.html



          https://www.gnu.org/software/bash/manual/html_node/Arrays.html






          share|improve this answer




















          • 2





            Thank you. I've revised a bit but feel free to make edits on it. I am not that good at guiding new users.

            – Weijun Zhou
            19 hours ago











          • thanks for the detailed explanation Weijun

            – Sunny
            5 hours ago















          4














          You are mixing the three different usages of parentheses in bash: array definition, command substitution, and command grouping. Command substitution and command grouping is standard syntax and all POSIX sh-like shells supports it. The array definition syntax is supported in shells that additionally have arrays.



          Array Definition




          Arrays are assigned to using compound assignments of the form



          name=(value1 value2 … )




          Your a=($x) creates an array with a single element, whose value is the value specified by $x, i.e. 1. When you echo an array by its name, in your case echo $a, it only echoes the first element of the array. So that's what you see, a 1 is printed to stdout.



          If you have multiple elements in an array you can use "$arrayname[*]" or "$arrayname[@]" to access them. The first combines all array elements into a single argument while the second put each element as an individual argument.



          This syntax is specific to bash and to other shells that allows for arrays in a similar way.



          Command substitution (POSIX)




          Command substitution allows the output of a command to replace the
          command itself. Command substitution occurs when a command is enclosed
          as follows:



          $(command)




          Your command $(echo foo) performs shell substitution and is parsed to foo. So it is the same as you have typed foo. Since the command foo does not exist, bash complains about it.



          Command grouping (POSIX)




          Bash provides two ways to group a list of commands to be executed as a
          unit. When commands are grouped, redirections may be applied to the
          entire command list. For example, the output of all the commands in
          the list may be redirected to a single stream.



          ( list )



          Placing a list of commands between parentheses causes a
          subshell environment to be created (see Command Execution
          Environment), and each of the commands in list to be executed in that
          subshell. Since the list is executed in a subshell, variable
          assignments do not remain in effect after the subshell completes.




          Your command (echo foo) executes echo foo in a subshell, so foo is echoed. This has nothing to do with the command substitution. As the manual says, variable assignments in subshell do not remain in effect after the subshell completes. This can come handy if you want to write simple one-liners. For example, instead of



          for l in 1..10; do mycommand $l; done; unset l


          You can write,



          ( for l in 1..10; do mycommand $l; done )


          Another helpful usage of a subshell is like using



          ( cd folder; ./mycommand )


          instead of



          cd folder; ./mycommand; cd -


          Your command echo ($x) falls into none of these three categories, and bash reports a syntax error.



          Quote your variables properly



          It is also worth mentioning that you should quote your variable expansions whenever applicable (for example, when a string should contain the value of a variable). See When is double-quoting necessary? for details.



          For example, the following would not invoke the particular constructs described above:



          x=1
          a="($x)"

          echo "$a"
          echo "($x)"


          Here, both calls to echo would output (1).



          Links to bash manual pages



          https://www.gnu.org/software/bash/manual/html_node/Command-Grouping.html



          https://www.gnu.org/software/bash/manual/html_node/Command-Substitution.html



          https://www.gnu.org/software/bash/manual/html_node/Arrays.html






          share|improve this answer




















          • 2





            Thank you. I've revised a bit but feel free to make edits on it. I am not that good at guiding new users.

            – Weijun Zhou
            19 hours ago











          • thanks for the detailed explanation Weijun

            – Sunny
            5 hours ago













          4












          4








          4







          You are mixing the three different usages of parentheses in bash: array definition, command substitution, and command grouping. Command substitution and command grouping is standard syntax and all POSIX sh-like shells supports it. The array definition syntax is supported in shells that additionally have arrays.



          Array Definition




          Arrays are assigned to using compound assignments of the form



          name=(value1 value2 … )




          Your a=($x) creates an array with a single element, whose value is the value specified by $x, i.e. 1. When you echo an array by its name, in your case echo $a, it only echoes the first element of the array. So that's what you see, a 1 is printed to stdout.



          If you have multiple elements in an array you can use "$arrayname[*]" or "$arrayname[@]" to access them. The first combines all array elements into a single argument while the second put each element as an individual argument.



          This syntax is specific to bash and to other shells that allows for arrays in a similar way.



          Command substitution (POSIX)




          Command substitution allows the output of a command to replace the
          command itself. Command substitution occurs when a command is enclosed
          as follows:



          $(command)




          Your command $(echo foo) performs shell substitution and is parsed to foo. So it is the same as you have typed foo. Since the command foo does not exist, bash complains about it.



          Command grouping (POSIX)




          Bash provides two ways to group a list of commands to be executed as a
          unit. When commands are grouped, redirections may be applied to the
          entire command list. For example, the output of all the commands in
          the list may be redirected to a single stream.



          ( list )



          Placing a list of commands between parentheses causes a
          subshell environment to be created (see Command Execution
          Environment), and each of the commands in list to be executed in that
          subshell. Since the list is executed in a subshell, variable
          assignments do not remain in effect after the subshell completes.




          Your command (echo foo) executes echo foo in a subshell, so foo is echoed. This has nothing to do with the command substitution. As the manual says, variable assignments in subshell do not remain in effect after the subshell completes. This can come handy if you want to write simple one-liners. For example, instead of



          for l in 1..10; do mycommand $l; done; unset l


          You can write,



          ( for l in 1..10; do mycommand $l; done )


          Another helpful usage of a subshell is like using



          ( cd folder; ./mycommand )


          instead of



          cd folder; ./mycommand; cd -


          Your command echo ($x) falls into none of these three categories, and bash reports a syntax error.



          Quote your variables properly



          It is also worth mentioning that you should quote your variable expansions whenever applicable (for example, when a string should contain the value of a variable). See When is double-quoting necessary? for details.



          For example, the following would not invoke the particular constructs described above:



          x=1
          a="($x)"

          echo "$a"
          echo "($x)"


          Here, both calls to echo would output (1).



          Links to bash manual pages



          https://www.gnu.org/software/bash/manual/html_node/Command-Grouping.html



          https://www.gnu.org/software/bash/manual/html_node/Command-Substitution.html



          https://www.gnu.org/software/bash/manual/html_node/Arrays.html






          share|improve this answer















          You are mixing the three different usages of parentheses in bash: array definition, command substitution, and command grouping. Command substitution and command grouping is standard syntax and all POSIX sh-like shells supports it. The array definition syntax is supported in shells that additionally have arrays.



          Array Definition




          Arrays are assigned to using compound assignments of the form



          name=(value1 value2 … )




          Your a=($x) creates an array with a single element, whose value is the value specified by $x, i.e. 1. When you echo an array by its name, in your case echo $a, it only echoes the first element of the array. So that's what you see, a 1 is printed to stdout.



          If you have multiple elements in an array you can use "$arrayname[*]" or "$arrayname[@]" to access them. The first combines all array elements into a single argument while the second put each element as an individual argument.



          This syntax is specific to bash and to other shells that allows for arrays in a similar way.



          Command substitution (POSIX)




          Command substitution allows the output of a command to replace the
          command itself. Command substitution occurs when a command is enclosed
          as follows:



          $(command)




          Your command $(echo foo) performs shell substitution and is parsed to foo. So it is the same as you have typed foo. Since the command foo does not exist, bash complains about it.



          Command grouping (POSIX)




          Bash provides two ways to group a list of commands to be executed as a
          unit. When commands are grouped, redirections may be applied to the
          entire command list. For example, the output of all the commands in
          the list may be redirected to a single stream.



          ( list )



          Placing a list of commands between parentheses causes a
          subshell environment to be created (see Command Execution
          Environment), and each of the commands in list to be executed in that
          subshell. Since the list is executed in a subshell, variable
          assignments do not remain in effect after the subshell completes.




          Your command (echo foo) executes echo foo in a subshell, so foo is echoed. This has nothing to do with the command substitution. As the manual says, variable assignments in subshell do not remain in effect after the subshell completes. This can come handy if you want to write simple one-liners. For example, instead of



          for l in 1..10; do mycommand $l; done; unset l


          You can write,



          ( for l in 1..10; do mycommand $l; done )


          Another helpful usage of a subshell is like using



          ( cd folder; ./mycommand )


          instead of



          cd folder; ./mycommand; cd -


          Your command echo ($x) falls into none of these three categories, and bash reports a syntax error.



          Quote your variables properly



          It is also worth mentioning that you should quote your variable expansions whenever applicable (for example, when a string should contain the value of a variable). See When is double-quoting necessary? for details.



          For example, the following would not invoke the particular constructs described above:



          x=1
          a="($x)"

          echo "$a"
          echo "($x)"


          Here, both calls to echo would output (1).



          Links to bash manual pages



          https://www.gnu.org/software/bash/manual/html_node/Command-Grouping.html



          https://www.gnu.org/software/bash/manual/html_node/Command-Substitution.html



          https://www.gnu.org/software/bash/manual/html_node/Arrays.html







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 14 hours ago

























          answered 19 hours ago









          Weijun ZhouWeijun Zhou

          1,661427




          1,661427







          • 2





            Thank you. I've revised a bit but feel free to make edits on it. I am not that good at guiding new users.

            – Weijun Zhou
            19 hours ago











          • thanks for the detailed explanation Weijun

            – Sunny
            5 hours ago












          • 2





            Thank you. I've revised a bit but feel free to make edits on it. I am not that good at guiding new users.

            – Weijun Zhou
            19 hours ago











          • thanks for the detailed explanation Weijun

            – Sunny
            5 hours ago







          2




          2





          Thank you. I've revised a bit but feel free to make edits on it. I am not that good at guiding new users.

          – Weijun Zhou
          19 hours ago





          Thank you. I've revised a bit but feel free to make edits on it. I am not that good at guiding new users.

          – Weijun Zhou
          19 hours ago













          thanks for the detailed explanation Weijun

          – Sunny
          5 hours ago





          thanks for the detailed explanation Weijun

          – Sunny
          5 hours ago










          Sunny is a new contributor. Be nice, and check out our Code of Conduct.









          draft saved

          draft discarded


















          Sunny is a new contributor. Be nice, and check out our Code of Conduct.












          Sunny is a new contributor. Be nice, and check out our Code of Conduct.











          Sunny is a new contributor. Be nice, and check out our Code of Conduct.














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