var=$[3 * 2] seem conflict with intuitive perception

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I am confused about the outer $ in var3=$[$var1 * $var2]



Suppose the following script:



$ var1=5; var2=6; var3=$[$var1 * $var2]; echo $var3
30


If $ is removed, it report error:



$ var1=5; var2=6; var3=[$var1 * $var2]; echo $var3
-bash: Algorithms: command not found
30
$ var=[3 * 2]; echo $var
-bash: Algorithms: command not found
[3*2]


I feel it very strange it to declare;



$ var=$[3 * 2]; echo $var
6


Perform very likely from intuitive perception:



$ var=$6; echo $var


It's odd.



What's the mechanism which force the syntax should do it this way?






bash






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    up vote
    0
    down vote

    favorite












    I am confused about the outer $ in var3=$[$var1 * $var2]



    Suppose the following script:



    $ var1=5; var2=6; var3=$[$var1 * $var2]; echo $var3
    30


    If $ is removed, it report error:



    $ var1=5; var2=6; var3=[$var1 * $var2]; echo $var3
    -bash: Algorithms: command not found
    30
    $ var=[3 * 2]; echo $var
    -bash: Algorithms: command not found
    [3*2]


    I feel it very strange it to declare;



    $ var=$[3 * 2]; echo $var
    6


    Perform very likely from intuitive perception:



    $ var=$6; echo $var


    It's odd.



    What's the mechanism which force the syntax should do it this way?






    bash






    share







    New contributor




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





















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I am confused about the outer $ in var3=$[$var1 * $var2]



      Suppose the following script:



      $ var1=5; var2=6; var3=$[$var1 * $var2]; echo $var3
      30


      If $ is removed, it report error:



      $ var1=5; var2=6; var3=[$var1 * $var2]; echo $var3
      -bash: Algorithms: command not found
      30
      $ var=[3 * 2]; echo $var
      -bash: Algorithms: command not found
      [3*2]


      I feel it very strange it to declare;



      $ var=$[3 * 2]; echo $var
      6


      Perform very likely from intuitive perception:



      $ var=$6; echo $var


      It's odd.



      What's the mechanism which force the syntax should do it this way?






      bash






      share







      New contributor




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











      I am confused about the outer $ in var3=$[$var1 * $var2]



      Suppose the following script:



      $ var1=5; var2=6; var3=$[$var1 * $var2]; echo $var3
      30


      If $ is removed, it report error:



      $ var1=5; var2=6; var3=[$var1 * $var2]; echo $var3
      -bash: Algorithms: command not found
      30
      $ var=[3 * 2]; echo $var
      -bash: Algorithms: command not found
      [3*2]


      I feel it very strange it to declare;



      $ var=$[3 * 2]; echo $var
      6


      Perform very likely from intuitive perception:



      $ var=$6; echo $var


      It's odd.



      What's the mechanism which force the syntax should do it this way?






      bash




      bash





      share







      New contributor




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










      share







      New contributor




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








      share



      share






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









      asked 2 mins ago









      rider dragon

      26526




      26526




      New contributor




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      New contributor





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






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

























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