AWK: Why $* works inside bash function but not under pipe?












3














@terdon in this post answered the related question of mine, but I missed one more question in that post.



Plz refer to the following commands:



calc(){ awk "BEGIN{ print $* }" ;}; calc "((3+(2^3)) * 34^2 / 9)-75.89"


The above commands work fine with calculated result '1337'.



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


But the above commands don't give any result while @terdon explained well about why.



Could you advise what made the first example work with $*?






awk







share|improve this question



























    3














    @terdon in this post answered the related question of mine, but I missed one more question in that post.



    Plz refer to the following commands:



    calc(){ awk "BEGIN{ print $* }" ;}; calc "((3+(2^3)) * 34^2 / 9)-75.89"


    The above commands work fine with calculated result '1337'.



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


    But the above commands don't give any result while @terdon explained well about why.



    Could you advise what made the first example work with $*?






    awk







    share|improve this question

























      3












      3








      3


      1





      @terdon in this post answered the related question of mine, but I missed one more question in that post.



      Plz refer to the following commands:



      calc(){ awk "BEGIN{ print $* }" ;}; calc "((3+(2^3)) * 34^2 / 9)-75.89"


      The above commands work fine with calculated result '1337'.



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


      But the above commands don't give any result while @terdon explained well about why.



      Could you advise what made the first example work with $*?






      awk







      share|improve this question













      @terdon in this post answered the related question of mine, but I missed one more question in that post.



      Plz refer to the following commands:



      calc(){ awk "BEGIN{ print $* }" ;}; calc "((3+(2^3)) * 34^2 / 9)-75.89"


      The above commands work fine with calculated result '1337'.



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


      But the above commands don't give any result while @terdon explained well about why.



      Could you advise what made the first example work with $*?






      awk




      awk






      share|improve this question













      share|improve this question











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      share|improve this question










      asked 6 hours ago









      user58029

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      505






















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          3














          $* refers to positional parameters - those variables which are referenced by $1 and $2 and so on, and are provided as arguments to scripts and functions. That's the key to your question.



          When you have interactive shell , there's no positional parameters set by default, so $* is empty. You can make it work if you set those via set "((3+(2^3)) * 34^2 / 9)-75.89" command, which will make $1 equal to that string.



          The difference with calc(){ awk "BEGIN{ print $* }" ;}; is that it's a function and functions can process positional parameters (theirs, not the shell's). When you call calc "((3+(2^3)) * 34^2 / 9)-75.89" you're calling a function with positional parameter "((3+(2^3)) * 34^2 / 9)-75.89". There $* won't be empty:



          $ calc(){ echo "Params: '$*'"; awk "BEGIN{ print $* }" ;}; calc "((3+(2^3)) * 34^2 / 9)-75.89"
          
Params: '((3+(2^3)) * 34^2 / 9)-75.89'
          
1337





          share|improve this answer























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            3














            $* refers to positional parameters - those variables which are referenced by $1 and $2 and so on, and are provided as arguments to scripts and functions. That's the key to your question.



            When you have interactive shell , there's no positional parameters set by default, so $* is empty. You can make it work if you set those via set "((3+(2^3)) * 34^2 / 9)-75.89" command, which will make $1 equal to that string.



            The difference with calc(){ awk "BEGIN{ print $* }" ;}; is that it's a function and functions can process positional parameters (theirs, not the shell's). When you call calc "((3+(2^3)) * 34^2 / 9)-75.89" you're calling a function with positional parameter "((3+(2^3)) * 34^2 / 9)-75.89". There $* won't be empty:



            $ calc(){ echo "Params: '$*'"; awk "BEGIN{ print $* }" ;}; calc "((3+(2^3)) * 34^2 / 9)-75.89"
            
Params: '((3+(2^3)) * 34^2 / 9)-75.89'
            
1337





            share|improve this answer




























              3














              $* refers to positional parameters - those variables which are referenced by $1 and $2 and so on, and are provided as arguments to scripts and functions. That's the key to your question.



              When you have interactive shell , there's no positional parameters set by default, so $* is empty. You can make it work if you set those via set "((3+(2^3)) * 34^2 / 9)-75.89" command, which will make $1 equal to that string.



              The difference with calc(){ awk "BEGIN{ print $* }" ;}; is that it's a function and functions can process positional parameters (theirs, not the shell's). When you call calc "((3+(2^3)) * 34^2 / 9)-75.89" you're calling a function with positional parameter "((3+(2^3)) * 34^2 / 9)-75.89". There $* won't be empty:



              $ calc(){ echo "Params: '$*'"; awk "BEGIN{ print $* }" ;}; calc "((3+(2^3)) * 34^2 / 9)-75.89"
              
Params: '((3+(2^3)) * 34^2 / 9)-75.89'
              
1337





              share|improve this answer


























                3












                3








                3






                $* refers to positional parameters - those variables which are referenced by $1 and $2 and so on, and are provided as arguments to scripts and functions. That's the key to your question.



                When you have interactive shell , there's no positional parameters set by default, so $* is empty. You can make it work if you set those via set "((3+(2^3)) * 34^2 / 9)-75.89" command, which will make $1 equal to that string.



                The difference with calc(){ awk "BEGIN{ print $* }" ;}; is that it's a function and functions can process positional parameters (theirs, not the shell's). When you call calc "((3+(2^3)) * 34^2 / 9)-75.89" you're calling a function with positional parameter "((3+(2^3)) * 34^2 / 9)-75.89". There $* won't be empty:



                $ calc(){ echo "Params: '$*'"; awk "BEGIN{ print $* }" ;}; calc "((3+(2^3)) * 34^2 / 9)-75.89"
                
Params: '((3+(2^3)) * 34^2 / 9)-75.89'
                
1337





                share|improve this answer














                $* refers to positional parameters - those variables which are referenced by $1 and $2 and so on, and are provided as arguments to scripts and functions. That's the key to your question.



                When you have interactive shell , there's no positional parameters set by default, so $* is empty. You can make it work if you set those via set "((3+(2^3)) * 34^2 / 9)-75.89" command, which will make $1 equal to that string.



                The difference with calc(){ awk "BEGIN{ print $* }" ;}; is that it's a function and functions can process positional parameters (theirs, not the shell's). When you call calc "((3+(2^3)) * 34^2 / 9)-75.89" you're calling a function with positional parameter "((3+(2^3)) * 34^2 / 9)-75.89". There $* won't be empty:



                $ calc(){ echo "Params: '$*'"; awk "BEGIN{ print $* }" ;}; calc "((3+(2^3)) * 34^2 / 9)-75.89"
                
Params: '((3+(2^3)) * 34^2 / 9)-75.89'
                
1337






                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 30 mins ago









                terdon♦

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                64.5k12136212










                answered 6 hours ago









                Sergiy Kolodyazhnyy

                69.6k9144306




                69.6k9144306






























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