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Rearrange c++ const and reference specifiers



The Next CEO of Stack OverflowCreating a regex to match html attributesSearch all mention of the current file's file nameIs there a way to convert a Vim regex literal to different magic modes?Recognizing expression written in assembly containing $ and new lineRegex to match any character including newlineNewline in sub-replace expressionBreak-down long strings and search/replace individual wordsloading back-reference groups “shortcut”regex find and replace all with capture groups throws E488: Trailing charactersUsing (neo)vim's regex to match up to but *excluding* a certain character?










1















I'm trying to convert



const std::string &s


to



std::string const & s


and can't seem to get it quite right. This is the regex I'm using:



%s/const (.-) &/1 const &


and the result I'm getting:



std::string const const std::string &s









share|improve this question







New contributor




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
























    1















    I'm trying to convert



    const std::string &s


    to



    std::string const & s


    and can't seem to get it quite right. This is the regex I'm using:



    %s/const (.-) &/1 const &


    and the result I'm getting:



    std::string const const std::string &s









    share|improve this question







    New contributor




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






















      1












      1








      1








      I'm trying to convert



      const std::string &s


      to



      std::string const & s


      and can't seem to get it quite right. This is the regex I'm using:



      %s/const (.-) &/1 const &


      and the result I'm getting:



      std::string const const std::string &s









      share|improve this question







      New contributor




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












      I'm trying to convert



      const std::string &s


      to



      std::string const & s


      and can't seem to get it quite right. This is the regex I'm using:



      %s/const (.-) &/1 const &


      and the result I'm getting:



      std::string const const std::string &s






      regular-expression c++






      share|improve this question







      New contributor




      Jay Mueller 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




      Jay Mueller 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




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









      asked 5 hours ago









      Jay MuellerJay Mueller

      82




      82




      New contributor




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





      New contributor





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






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




















          1 Answer
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          active

          oldest

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          3














          Just escape the & with a backslash.



          %s/const (.-) &/1 const &


          The & in the replace part inserts the entire matched pattern.



          See :help s/&.






          share|improve this answer























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            1 Answer
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            1 Answer
            1






            active

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            active

            oldest

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            active

            oldest

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            3














            Just escape the & with a backslash.



            %s/const (.-) &/1 const &


            The & in the replace part inserts the entire matched pattern.



            See :help s/&.






            share|improve this answer



























              3














              Just escape the & with a backslash.



              %s/const (.-) &/1 const &


              The & in the replace part inserts the entire matched pattern.



              See :help s/&.






              share|improve this answer

























                3












                3








                3







                Just escape the & with a backslash.



                %s/const (.-) &/1 const &


                The & in the replace part inserts the entire matched pattern.



                See :help s/&.






                share|improve this answer













                Just escape the & with a backslash.



                %s/const (.-) &/1 const &


                The & in the replace part inserts the entire matched pattern.



                See :help s/&.







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 5 hours ago









                RalfRalf

                3,1551317




                3,1551317




















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









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














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