How do I use command-line arguments to `sh` in the `-c` command string?












3
















I know that



sh -c 'echo $1' sh 4


will output 4.
and



sh -c 'echo $2' sh 4 5


will output 5.



But I cannot understand how the parameters after the second sh were passed to the command next to sh -c. I read the man page of both bash and dash but could not find the introduction about this kind of syntax.






command-line bash sh dash-shell







share|improve this question





























    3
















    I know that



    sh -c 'echo $1' sh 4


    will output 4.
    and



    sh -c 'echo $2' sh 4 5


    will output 5.



    But I cannot understand how the parameters after the second sh were passed to the command next to sh -c. I read the man page of both bash and dash but could not find the introduction about this kind of syntax.






    command-line bash sh dash-shell







    share|improve this question



























      3












      3








      3









      I know that



      sh -c 'echo $1' sh 4


      will output 4.
      and



      sh -c 'echo $2' sh 4 5


      will output 5.



      But I cannot understand how the parameters after the second sh were passed to the command next to sh -c. I read the man page of both bash and dash but could not find the introduction about this kind of syntax.






      command-line bash sh dash-shell







      share|improve this question

















      I know that



      sh -c 'echo $1' sh 4


      will output 4.
      and



      sh -c 'echo $2' sh 4 5


      will output 5.



      But I cannot understand how the parameters after the second sh were passed to the command next to sh -c. I read the man page of both bash and dash but could not find the introduction about this kind of syntax.






      command-line bash sh dash-shell




      command-line bash sh dash-shell






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited yesterday









      dessert

      22.2k56198




      22.2k56198










      asked yesterday









      gbcatgbcat

      185




      185






















          2 Answers
          2






          active

          oldest

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          4














          From man sh:



          -c string If  the  -c  option  is  present, then commands are read from
          
          string.  If there are arguments after the string,  they  are
          
          assigned to the positional parameters, starting with $0.


          In your command the second sh is just a parameter with position 0 while 4 has position 1 and so on.



          You can run this to check:



          $ sh -c 'echo $0' sh 4 5
          
sh





          share|improve this answer























          • That seems to be the text in Bash's man page. It's actually incorrect in that it implies that $0 is one of the positional parameters. It isn't. The POSIX definition states that "A positional parameter is a parameter denoted by the decimal value represented by one or more digits, other than the single digit 0.", and Bash's manual itself contains that same definition.
            – ilkkachu
            yesterday





















          4














          This behavior is actually specified by POSIX standard, which all Bourne-like shells should support to claim themselves portable.




          sh -c [-abCefhimnuvx] [-o option]... [+abCefhimnuvx] [+o option]...
          command_string [command_name [argument...]]




          See the command_string parameter ? Now let's look at -c flag description:




          -c



          Read commands from the command_string operand. Set the value of special parameter 0 (see Special Parameters) from the value of the command_name operand and the positional parameters ($1, $2, and so on) in sequence from the remaining argument operands. No commands shall be read from the standard input.




          In other words, where in normal shell script $0 (which is usually shell name in interactive mode or script name when you run a script) would be set by the shell itself, with -c you have to specify it yourself. Thus,



          sh -c 'echo Hi, I am $0 , my first positional parameter is $1' foobar 5


          would set the process name to sh foobar.



          Just in case you're wondering what $0 is, it's also covered in "Special Parameters" section of Shell Command Language Specifications:




          0



          (Zero.) Expands to the name of the shell or shell script.







          share|improve this answer























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            2 Answers
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            2 Answers
            2






            active

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            active

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            active

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            4














            From man sh:



            -c string If  the  -c  option  is  present, then commands are read from
            
          string.  If there are arguments after the string,  they  are
            
          assigned to the positional parameters, starting with $0.


            In your command the second sh is just a parameter with position 0 while 4 has position 1 and so on.



            You can run this to check:



            $ sh -c 'echo $0' sh 4 5
            
sh





            share|improve this answer























            • That seems to be the text in Bash's man page. It's actually incorrect in that it implies that $0 is one of the positional parameters. It isn't. The POSIX definition states that "A positional parameter is a parameter denoted by the decimal value represented by one or more digits, other than the single digit 0.", and Bash's manual itself contains that same definition.
              – ilkkachu
              yesterday


















            4














            From man sh:



            -c string If  the  -c  option  is  present, then commands are read from
            
          string.  If there are arguments after the string,  they  are
            
          assigned to the positional parameters, starting with $0.


            In your command the second sh is just a parameter with position 0 while 4 has position 1 and so on.



            You can run this to check:



            $ sh -c 'echo $0' sh 4 5
            
sh





            share|improve this answer























            • That seems to be the text in Bash's man page. It's actually incorrect in that it implies that $0 is one of the positional parameters. It isn't. The POSIX definition states that "A positional parameter is a parameter denoted by the decimal value represented by one or more digits, other than the single digit 0.", and Bash's manual itself contains that same definition.
              – ilkkachu
              yesterday
















            4












            4








            4






            From man sh:



            -c string If  the  -c  option  is  present, then commands are read from
            
          string.  If there are arguments after the string,  they  are
            
          assigned to the positional parameters, starting with $0.


            In your command the second sh is just a parameter with position 0 while 4 has position 1 and so on.



            You can run this to check:



            $ sh -c 'echo $0' sh 4 5
            
sh





            share|improve this answer














            From man sh:



            -c string If  the  -c  option  is  present, then commands are read from
            
          string.  If there are arguments after the string,  they  are
            
          assigned to the positional parameters, starting with $0.


            In your command the second sh is just a parameter with position 0 while 4 has position 1 and so on.



            You can run this to check:



            $ sh -c 'echo $0' sh 4 5
            
sh






            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited yesterday









            dessert

            22.2k56198




            22.2k56198










            answered yesterday









            eyadofeyadof

            1,20411517




            1,20411517












            • That seems to be the text in Bash's man page. It's actually incorrect in that it implies that $0 is one of the positional parameters. It isn't. The POSIX definition states that "A positional parameter is a parameter denoted by the decimal value represented by one or more digits, other than the single digit 0.", and Bash's manual itself contains that same definition.
              – ilkkachu
              yesterday




















            • That seems to be the text in Bash's man page. It's actually incorrect in that it implies that $0 is one of the positional parameters. It isn't. The POSIX definition states that "A positional parameter is a parameter denoted by the decimal value represented by one or more digits, other than the single digit 0.", and Bash's manual itself contains that same definition.
              – ilkkachu
              yesterday


















            That seems to be the text in Bash's man page. It's actually incorrect in that it implies that $0 is one of the positional parameters. It isn't. The POSIX definition states that "A positional parameter is a parameter denoted by the decimal value represented by one or more digits, other than the single digit 0.", and Bash's manual itself contains that same definition.
            – ilkkachu
            yesterday






            That seems to be the text in Bash's man page. It's actually incorrect in that it implies that $0 is one of the positional parameters. It isn't. The POSIX definition states that "A positional parameter is a parameter denoted by the decimal value represented by one or more digits, other than the single digit 0.", and Bash's manual itself contains that same definition.
            – ilkkachu
            yesterday















            4














            This behavior is actually specified by POSIX standard, which all Bourne-like shells should support to claim themselves portable.




            sh -c [-abCefhimnuvx] [-o option]... [+abCefhimnuvx] [+o option]...
            command_string [command_name [argument...]]




            See the command_string parameter ? Now let's look at -c flag description:




            -c



            Read commands from the command_string operand. Set the value of special parameter 0 (see Special Parameters) from the value of the command_name operand and the positional parameters ($1, $2, and so on) in sequence from the remaining argument operands. No commands shall be read from the standard input.




            In other words, where in normal shell script $0 (which is usually shell name in interactive mode or script name when you run a script) would be set by the shell itself, with -c you have to specify it yourself. Thus,



            sh -c 'echo Hi, I am $0 , my first positional parameter is $1' foobar 5


            would set the process name to sh foobar.



            Just in case you're wondering what $0 is, it's also covered in "Special Parameters" section of Shell Command Language Specifications:




            0



            (Zero.) Expands to the name of the shell or shell script.







            share|improve this answer




























              4














              This behavior is actually specified by POSIX standard, which all Bourne-like shells should support to claim themselves portable.




              sh -c [-abCefhimnuvx] [-o option]... [+abCefhimnuvx] [+o option]...
              command_string [command_name [argument...]]




              See the command_string parameter ? Now let's look at -c flag description:




              -c



              Read commands from the command_string operand. Set the value of special parameter 0 (see Special Parameters) from the value of the command_name operand and the positional parameters ($1, $2, and so on) in sequence from the remaining argument operands. No commands shall be read from the standard input.




              In other words, where in normal shell script $0 (which is usually shell name in interactive mode or script name when you run a script) would be set by the shell itself, with -c you have to specify it yourself. Thus,



              sh -c 'echo Hi, I am $0 , my first positional parameter is $1' foobar 5


              would set the process name to sh foobar.



              Just in case you're wondering what $0 is, it's also covered in "Special Parameters" section of Shell Command Language Specifications:




              0



              (Zero.) Expands to the name of the shell or shell script.







              share|improve this answer


























                4












                4








                4






                This behavior is actually specified by POSIX standard, which all Bourne-like shells should support to claim themselves portable.




                sh -c [-abCefhimnuvx] [-o option]... [+abCefhimnuvx] [+o option]...
                command_string [command_name [argument...]]




                See the command_string parameter ? Now let's look at -c flag description:




                -c



                Read commands from the command_string operand. Set the value of special parameter 0 (see Special Parameters) from the value of the command_name operand and the positional parameters ($1, $2, and so on) in sequence from the remaining argument operands. No commands shall be read from the standard input.




                In other words, where in normal shell script $0 (which is usually shell name in interactive mode or script name when you run a script) would be set by the shell itself, with -c you have to specify it yourself. Thus,



                sh -c 'echo Hi, I am $0 , my first positional parameter is $1' foobar 5


                would set the process name to sh foobar.



                Just in case you're wondering what $0 is, it's also covered in "Special Parameters" section of Shell Command Language Specifications:




                0



                (Zero.) Expands to the name of the shell or shell script.







                share|improve this answer














                This behavior is actually specified by POSIX standard, which all Bourne-like shells should support to claim themselves portable.




                sh -c [-abCefhimnuvx] [-o option]... [+abCefhimnuvx] [+o option]...
                command_string [command_name [argument...]]




                See the command_string parameter ? Now let's look at -c flag description:




                -c



                Read commands from the command_string operand. Set the value of special parameter 0 (see Special Parameters) from the value of the command_name operand and the positional parameters ($1, $2, and so on) in sequence from the remaining argument operands. No commands shall be read from the standard input.




                In other words, where in normal shell script $0 (which is usually shell name in interactive mode or script name when you run a script) would be set by the shell itself, with -c you have to specify it yourself. Thus,



                sh -c 'echo Hi, I am $0 , my first positional parameter is $1' foobar 5


                would set the process name to sh foobar.



                Just in case you're wondering what $0 is, it's also covered in "Special Parameters" section of Shell Command Language Specifications:




                0



                (Zero.) Expands to the name of the shell or shell script.








                share|improve this answer














                share|improve this answer



                share|improve this answer








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                Sergiy KolodyazhnyySergiy Kolodyazhnyy

                70.1k9144307




                70.1k9144307






























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