AWK: why does <(cat) work for stdin, but $* doesn't?












3














echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN{ print $(cat) }"


The above syntax works fine with the calculated result '1337'.



echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN{ print $* }"


But the above syntax doesn't work, though there's no error.



Plz advise.










share|improve this question
























  • What are you expecting it to do? And do you want to ask about $(cat) or <(cat)? They are two very different things.
    – terdon
    53 mins ago


















3














echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN{ print $(cat) }"


The above syntax works fine with the calculated result '1337'.



echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN{ print $* }"


But the above syntax doesn't work, though there's no error.



Plz advise.










share|improve this question
























  • What are you expecting it to do? And do you want to ask about $(cat) or <(cat)? They are two very different things.
    – terdon
    53 mins ago
















3












3








3


1





echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN{ print $(cat) }"


The above syntax works fine with the calculated result '1337'.



echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN{ print $* }"


But the above syntax doesn't work, though there's no error.



Plz advise.










share|improve this question















echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN{ print $(cat) }"


The above syntax works fine with the calculated result '1337'.



echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN{ print $* }"


But the above syntax doesn't work, though there's no error.



Plz advise.







awk






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 53 mins ago









terdon

64.4k12136212




64.4k12136212










asked 1 hour ago









user58029

184




184












  • What are you expecting it to do? And do you want to ask about $(cat) or <(cat)? They are two very different things.
    – terdon
    53 mins ago




















  • What are you expecting it to do? And do you want to ask about $(cat) or <(cat)? They are two very different things.
    – terdon
    53 mins ago


















What are you expecting it to do? And do you want to ask about $(cat) or <(cat)? They are two very different things.
– terdon
53 mins ago






What are you expecting it to do? And do you want to ask about $(cat) or <(cat)? They are two very different things.
– terdon
53 mins ago












1 Answer
1






active

oldest

votes


















3














The $(command) syntax will return the output of command. Here, you are using the very simple cat program whose only job is to copy everything from standard input (stdin) into standard output (stdout). Since you are running the awk script inside double quotes, the $(cat) is expanded by the shell before the awk script is run, so it reads the echo output into its stdin and duly copies it to its stdout. This is then passed to the awk script. You can see this in action with set -x:



$ set -x
$ echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN{ print $(cat) }"
+ echo '((3+(2^3)) * 34^2 / 9)-75.89'
++ cat
+ awk 'BEGIN{ print ((3+(2^3)) * 34^2 / 9)-75.89 }'
1337


So, awk is actually running BEGIN{ print ((3+(2^3)) * 34^2 / 9)-75.89 }' which returns 1337.



Now, the $* is a special shell variable that expands to all the positional parameters given to a shell script (see man bash):



   *      Expands to the positional parameters, starting from one.  When the expan‐
sion is not within double quotes, each positional parameter expands to a
separate word. In contexts where it is performed, those words are sub‐
ject to further word splitting and pathname expansion. When the expan‐
sion occurs within double quotes, it expands to a single word with the
value of each parameter separated by the first character of the IFS spe‐
cial variable. That is, "$*" is equivalent to "$1c$2c...", where c is
the first character of the value of the IFS variable. If IFS is unset,
the parameters are separated by spaces. If IFS is null, the parameters
are joined without intervening separators.


However, this variable is empty here since there is no shell script and no parameters. Therefore, the awk script becomes:



$ echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN{ print $* }"
+ awk 'BEGIN{ print }'
+ echo '((3+(2^3)) * 34^2 / 9)-75.89'


The $* expands to an empty string, and awk is told to print an empty string, and this is why you get no output.





You might want to just use bc instead:



$ echo '((3+(2^3)) * 34^2 / 9)-75.89' | bc
1336.11





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    1 Answer
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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3














    The $(command) syntax will return the output of command. Here, you are using the very simple cat program whose only job is to copy everything from standard input (stdin) into standard output (stdout). Since you are running the awk script inside double quotes, the $(cat) is expanded by the shell before the awk script is run, so it reads the echo output into its stdin and duly copies it to its stdout. This is then passed to the awk script. You can see this in action with set -x:



    $ set -x
    $ echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN{ print $(cat) }"
    + echo '((3+(2^3)) * 34^2 / 9)-75.89'
    ++ cat
    + awk 'BEGIN{ print ((3+(2^3)) * 34^2 / 9)-75.89 }'
    1337


    So, awk is actually running BEGIN{ print ((3+(2^3)) * 34^2 / 9)-75.89 }' which returns 1337.



    Now, the $* is a special shell variable that expands to all the positional parameters given to a shell script (see man bash):



       *      Expands to the positional parameters, starting from one.  When the expan‐
    sion is not within double quotes, each positional parameter expands to a
    separate word. In contexts where it is performed, those words are sub‐
    ject to further word splitting and pathname expansion. When the expan‐
    sion occurs within double quotes, it expands to a single word with the
    value of each parameter separated by the first character of the IFS spe‐
    cial variable. That is, "$*" is equivalent to "$1c$2c...", where c is
    the first character of the value of the IFS variable. If IFS is unset,
    the parameters are separated by spaces. If IFS is null, the parameters
    are joined without intervening separators.


    However, this variable is empty here since there is no shell script and no parameters. Therefore, the awk script becomes:



    $ echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN{ print $* }"
    + awk 'BEGIN{ print }'
    + echo '((3+(2^3)) * 34^2 / 9)-75.89'


    The $* expands to an empty string, and awk is told to print an empty string, and this is why you get no output.





    You might want to just use bc instead:



    $ echo '((3+(2^3)) * 34^2 / 9)-75.89' | bc
    1336.11





    share|improve this answer




























      3














      The $(command) syntax will return the output of command. Here, you are using the very simple cat program whose only job is to copy everything from standard input (stdin) into standard output (stdout). Since you are running the awk script inside double quotes, the $(cat) is expanded by the shell before the awk script is run, so it reads the echo output into its stdin and duly copies it to its stdout. This is then passed to the awk script. You can see this in action with set -x:



      $ set -x
      $ echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN{ print $(cat) }"
      + echo '((3+(2^3)) * 34^2 / 9)-75.89'
      ++ cat
      + awk 'BEGIN{ print ((3+(2^3)) * 34^2 / 9)-75.89 }'
      1337


      So, awk is actually running BEGIN{ print ((3+(2^3)) * 34^2 / 9)-75.89 }' which returns 1337.



      Now, the $* is a special shell variable that expands to all the positional parameters given to a shell script (see man bash):



         *      Expands to the positional parameters, starting from one.  When the expan‐
      sion is not within double quotes, each positional parameter expands to a
      separate word. In contexts where it is performed, those words are sub‐
      ject to further word splitting and pathname expansion. When the expan‐
      sion occurs within double quotes, it expands to a single word with the
      value of each parameter separated by the first character of the IFS spe‐
      cial variable. That is, "$*" is equivalent to "$1c$2c...", where c is
      the first character of the value of the IFS variable. If IFS is unset,
      the parameters are separated by spaces. If IFS is null, the parameters
      are joined without intervening separators.


      However, this variable is empty here since there is no shell script and no parameters. Therefore, the awk script becomes:



      $ echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN{ print $* }"
      + awk 'BEGIN{ print }'
      + echo '((3+(2^3)) * 34^2 / 9)-75.89'


      The $* expands to an empty string, and awk is told to print an empty string, and this is why you get no output.





      You might want to just use bc instead:



      $ echo '((3+(2^3)) * 34^2 / 9)-75.89' | bc
      1336.11





      share|improve this answer


























        3












        3








        3






        The $(command) syntax will return the output of command. Here, you are using the very simple cat program whose only job is to copy everything from standard input (stdin) into standard output (stdout). Since you are running the awk script inside double quotes, the $(cat) is expanded by the shell before the awk script is run, so it reads the echo output into its stdin and duly copies it to its stdout. This is then passed to the awk script. You can see this in action with set -x:



        $ set -x
        $ echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN{ print $(cat) }"
        + echo '((3+(2^3)) * 34^2 / 9)-75.89'
        ++ cat
        + awk 'BEGIN{ print ((3+(2^3)) * 34^2 / 9)-75.89 }'
        1337


        So, awk is actually running BEGIN{ print ((3+(2^3)) * 34^2 / 9)-75.89 }' which returns 1337.



        Now, the $* is a special shell variable that expands to all the positional parameters given to a shell script (see man bash):



           *      Expands to the positional parameters, starting from one.  When the expan‐
        sion is not within double quotes, each positional parameter expands to a
        separate word. In contexts where it is performed, those words are sub‐
        ject to further word splitting and pathname expansion. When the expan‐
        sion occurs within double quotes, it expands to a single word with the
        value of each parameter separated by the first character of the IFS spe‐
        cial variable. That is, "$*" is equivalent to "$1c$2c...", where c is
        the first character of the value of the IFS variable. If IFS is unset,
        the parameters are separated by spaces. If IFS is null, the parameters
        are joined without intervening separators.


        However, this variable is empty here since there is no shell script and no parameters. Therefore, the awk script becomes:



        $ echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN{ print $* }"
        + awk 'BEGIN{ print }'
        + echo '((3+(2^3)) * 34^2 / 9)-75.89'


        The $* expands to an empty string, and awk is told to print an empty string, and this is why you get no output.





        You might want to just use bc instead:



        $ echo '((3+(2^3)) * 34^2 / 9)-75.89' | bc
        1336.11





        share|improve this answer














        The $(command) syntax will return the output of command. Here, you are using the very simple cat program whose only job is to copy everything from standard input (stdin) into standard output (stdout). Since you are running the awk script inside double quotes, the $(cat) is expanded by the shell before the awk script is run, so it reads the echo output into its stdin and duly copies it to its stdout. This is then passed to the awk script. You can see this in action with set -x:



        $ set -x
        $ echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN{ print $(cat) }"
        + echo '((3+(2^3)) * 34^2 / 9)-75.89'
        ++ cat
        + awk 'BEGIN{ print ((3+(2^3)) * 34^2 / 9)-75.89 }'
        1337


        So, awk is actually running BEGIN{ print ((3+(2^3)) * 34^2 / 9)-75.89 }' which returns 1337.



        Now, the $* is a special shell variable that expands to all the positional parameters given to a shell script (see man bash):



           *      Expands to the positional parameters, starting from one.  When the expan‐
        sion is not within double quotes, each positional parameter expands to a
        separate word. In contexts where it is performed, those words are sub‐
        ject to further word splitting and pathname expansion. When the expan‐
        sion occurs within double quotes, it expands to a single word with the
        value of each parameter separated by the first character of the IFS spe‐
        cial variable. That is, "$*" is equivalent to "$1c$2c...", where c is
        the first character of the value of the IFS variable. If IFS is unset,
        the parameters are separated by spaces. If IFS is null, the parameters
        are joined without intervening separators.


        However, this variable is empty here since there is no shell script and no parameters. Therefore, the awk script becomes:



        $ echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN{ print $* }"
        + awk 'BEGIN{ print }'
        + echo '((3+(2^3)) * 34^2 / 9)-75.89'


        The $* expands to an empty string, and awk is told to print an empty string, and this is why you get no output.





        You might want to just use bc instead:



        $ echo '((3+(2^3)) * 34^2 / 9)-75.89' | bc
        1336.11






        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 40 mins ago

























        answered 46 mins ago









        terdon

        64.4k12136212




        64.4k12136212






























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