Standard Behavior Of An Empty Macro Preceding A Preprocessing Directive












0














Take, for example, the following:



#define FOO

FOO #define BAR 1

BAR


What should, according to each of the ANSI C and C99 standards, be the preprocessed output of the above code?



It seems to me that this should be evaluated to 1; however, running the above example through both gcc -E and clang -E produces the following:



    #define BAR 1

BAR





c c-preprocessor preprocessor







share|improve this question






















  • # should be the first non-space character in a line with directive, otherwise it gets ignored
    – qrdl
    Nov 18 '18 at 8:00












  • @qrdl: it does not get ignored: it will produce an error because # is not a valid token after preprocessing.
    – rici
    Nov 18 '18 at 8:08










  • @rici Obviously compiler will complain about it, but OP is talking just about CPP here, and CPP ignores it.
    – qrdl
    Nov 18 '18 at 8:11
















0














Take, for example, the following:



#define FOO

FOO #define BAR 1

BAR


What should, according to each of the ANSI C and C99 standards, be the preprocessed output of the above code?



It seems to me that this should be evaluated to 1; however, running the above example through both gcc -E and clang -E produces the following:



    #define BAR 1

BAR





c c-preprocessor preprocessor







share|improve this question






















  • # should be the first non-space character in a line with directive, otherwise it gets ignored
    – qrdl
    Nov 18 '18 at 8:00












  • @qrdl: it does not get ignored: it will produce an error because # is not a valid token after preprocessing.
    – rici
    Nov 18 '18 at 8:08










  • @rici Obviously compiler will complain about it, but OP is talking just about CPP here, and CPP ignores it.
    – qrdl
    Nov 18 '18 at 8:11














0












0








0







Take, for example, the following:



#define FOO

FOO #define BAR 1

BAR


What should, according to each of the ANSI C and C99 standards, be the preprocessed output of the above code?



It seems to me that this should be evaluated to 1; however, running the above example through both gcc -E and clang -E produces the following:



    #define BAR 1

BAR





c c-preprocessor preprocessor







share|improve this question













Take, for example, the following:



#define FOO

FOO #define BAR 1

BAR


What should, according to each of the ANSI C and C99 standards, be the preprocessed output of the above code?



It seems to me that this should be evaluated to 1; however, running the above example through both gcc -E and clang -E produces the following:



    #define BAR 1

BAR





c c-preprocessor preprocessor




c c-preprocessor preprocessor






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 18 '18 at 7:50









jinscoe123jinscoe123

17912




17912












  • # should be the first non-space character in a line with directive, otherwise it gets ignored
    – qrdl
    Nov 18 '18 at 8:00












  • @qrdl: it does not get ignored: it will produce an error because # is not a valid token after preprocessing.
    – rici
    Nov 18 '18 at 8:08










  • @rici Obviously compiler will complain about it, but OP is talking just about CPP here, and CPP ignores it.
    – qrdl
    Nov 18 '18 at 8:11


















  • # should be the first non-space character in a line with directive, otherwise it gets ignored
    – qrdl
    Nov 18 '18 at 8:00












  • @qrdl: it does not get ignored: it will produce an error because # is not a valid token after preprocessing.
    – rici
    Nov 18 '18 at 8:08










  • @rici Obviously compiler will complain about it, but OP is talking just about CPP here, and CPP ignores it.
    – qrdl
    Nov 18 '18 at 8:11
















# should be the first non-space character in a line with directive, otherwise it gets ignored
– qrdl
Nov 18 '18 at 8:00






# should be the first non-space character in a line with directive, otherwise it gets ignored
– qrdl
Nov 18 '18 at 8:00














@qrdl: it does not get ignored: it will produce an error because # is not a valid token after preprocessing.
– rici
Nov 18 '18 at 8:08




@qrdl: it does not get ignored: it will produce an error because # is not a valid token after preprocessing.
– rici
Nov 18 '18 at 8:08












@rici Obviously compiler will complain about it, but OP is talking just about CPP here, and CPP ignores it.
– qrdl
Nov 18 '18 at 8:11




@rici Obviously compiler will complain about it, but OP is talking just about CPP here, and CPP ignores it.
– qrdl
Nov 18 '18 at 8:11












3 Answers
3






active

oldest

votes


















1














The draft standard "ISO/IEC 9899:201x Committee Draft — April 12, 2011 N1570" section 6.10 actually contains an example of this:




EXAMPLE In: 



#define EMPTY 

EMPTY # include <file.h>


the sequence of preprocessing tokens on the second line is not a preprocessing directive, because it does not begin with a # at the start of translation phase 4, even though it will do so after the macro EMPTY has been replaced.




It tells us that "... the second line is not a preprocessing directive ..."



So for your code



FOO #define BAR 1


is not a preprocessing directive meaning that only FOO will be replaced and BAR will not be defined. Consequently the output of the preprocessor is:



 #define BAR 1

BAR





share|improve this answer





























    1














    Your code is not valid



    ISO/IEC 9899:2011, Section 6.10 Preprocessing directives:




    A preprocessing directive consists of a sequence of preprocessing
    tokens that satisfies the following constraints: The first token in
    the sequence is a # preprocessing token     that (at the start of
    translation phase 4) is either the first character in the source file
    (optionally after white space containing no new-line characters) or
    that follows white space containing at least one new-line character.







    share|improve this answer





























      0














      This example actually occurs in the Standard (C17 6.10/8):




      EXAMPLE In: 



      #define EMPTY
      
EMPTY # include <file.h>


      the sequence of preprocessing tokens on the second line is not a preprocessing directive, because it does not begin with a # at the start of translation phase 4, even though it will do so after the macro EMPTY has been replaced.




      So the output you see from gcc -E is correct. (Note: the amount of whitespace here is not significant , at that stage of translation the program has been translated to a sequence of preprocessing tokens; the different amounts of whitespace in the output is just an artefact of how gcc -E works).






      share|improve this answer





















      • Thank you as well, but, unfortunately, I can only accept one answer, and 4386427 was too quick. :)
        – jinscoe123
        Nov 18 '18 at 8:39











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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1














      The draft standard "ISO/IEC 9899:201x Committee Draft — April 12, 2011 N1570" section 6.10 actually contains an example of this:




      EXAMPLE In: 



      #define EMPTY 
      
EMPTY # include <file.h>


      the sequence of preprocessing tokens on the second line is not a preprocessing directive, because it does not begin with a # at the start of translation phase 4, even though it will do so after the macro EMPTY has been replaced.




      It tells us that "... the second line is not a preprocessing directive ..."



      So for your code



      FOO #define BAR 1


      is not a preprocessing directive meaning that only FOO will be replaced and BAR will not be defined. Consequently the output of the preprocessor is:



       #define BAR 1
      
BAR





      share|improve this answer


























        1














        The draft standard "ISO/IEC 9899:201x Committee Draft — April 12, 2011 N1570" section 6.10 actually contains an example of this:




        EXAMPLE In: 



        #define EMPTY 
        
EMPTY # include <file.h>


        the sequence of preprocessing tokens on the second line is not a preprocessing directive, because it does not begin with a # at the start of translation phase 4, even though it will do so after the macro EMPTY has been replaced.




        It tells us that "... the second line is not a preprocessing directive ..."



        So for your code



        FOO #define BAR 1


        is not a preprocessing directive meaning that only FOO will be replaced and BAR will not be defined. Consequently the output of the preprocessor is:



         #define BAR 1
        
BAR





        share|improve this answer
























          1












          1








          1






          The draft standard "ISO/IEC 9899:201x Committee Draft — April 12, 2011 N1570" section 6.10 actually contains an example of this:




          EXAMPLE In: 



          #define EMPTY 
          
EMPTY # include <file.h>


          the sequence of preprocessing tokens on the second line is not a preprocessing directive, because it does not begin with a # at the start of translation phase 4, even though it will do so after the macro EMPTY has been replaced.




          It tells us that "... the second line is not a preprocessing directive ..."



          So for your code



          FOO #define BAR 1


          is not a preprocessing directive meaning that only FOO will be replaced and BAR will not be defined. Consequently the output of the preprocessor is:



           #define BAR 1
          
BAR





          share|improve this answer












          The draft standard "ISO/IEC 9899:201x Committee Draft — April 12, 2011 N1570" section 6.10 actually contains an example of this:




          EXAMPLE In: 



          #define EMPTY 
          
EMPTY # include <file.h>


          the sequence of preprocessing tokens on the second line is not a preprocessing directive, because it does not begin with a # at the start of translation phase 4, even though it will do so after the macro EMPTY has been replaced.




          It tells us that "... the second line is not a preprocessing directive ..."



          So for your code



          FOO #define BAR 1


          is not a preprocessing directive meaning that only FOO will be replaced and BAR will not be defined. Consequently the output of the preprocessor is:



           #define BAR 1
          
BAR






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 18 '18 at 8:30









          43864274386427

          20.5k31845




          20.5k31845

























              1














              Your code is not valid



              ISO/IEC 9899:2011, Section 6.10 Preprocessing directives:




              A preprocessing directive consists of a sequence of preprocessing
              tokens that satisfies the following constraints: The first token in
              the sequence is a # preprocessing token               that (at the start of
              translation phase 4) is either the first character in the source file
              (optionally after white space containing no new-line characters) or
              that follows white space containing at least one new-line character.







              share|improve this answer


























                1














                Your code is not valid



                ISO/IEC 9899:2011, Section 6.10 Preprocessing directives:




                A preprocessing directive consists of a sequence of preprocessing
                tokens that satisfies the following constraints: The first token in
                the sequence is a # preprocessing token                 that (at the start of
                translation phase 4) is either the first character in the source file
                (optionally after white space containing no new-line characters) or
                that follows white space containing at least one new-line character.







                share|improve this answer
























                  1












                  1








                  1






                  Your code is not valid



                  ISO/IEC 9899:2011, Section 6.10 Preprocessing directives:




                  A preprocessing directive consists of a sequence of preprocessing
                  tokens that satisfies the following constraints: The first token in
                  the sequence is a # preprocessing token                   that (at the start of
                  translation phase 4) is either the first character in the source file
                  (optionally after white space containing no new-line characters) or
                  that follows white space containing at least one new-line character.







                  share|improve this answer












                  Your code is not valid



                  ISO/IEC 9899:2011, Section 6.10 Preprocessing directives:




                  A preprocessing directive consists of a sequence of preprocessing
                  tokens that satisfies the following constraints: The first token in
                  the sequence is a # preprocessing token                   that (at the start of
                  translation phase 4) is either the first character in the source file
                  (optionally after white space containing no new-line characters) or
                  that follows white space containing at least one new-line character.








                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Nov 18 '18 at 8:01









                  Keine LustKeine Lust

                  26.8k43363




                  26.8k43363























                      0














                      This example actually occurs in the Standard (C17 6.10/8):




                      EXAMPLE In: 



                      #define EMPTY
                      
EMPTY # include <file.h>


                      the sequence of preprocessing tokens on the second line is not a preprocessing directive, because it does not begin with a # at the start of translation phase 4, even though it will do so after the macro EMPTY has been replaced.




                      So the output you see from gcc -E is correct. (Note: the amount of whitespace here is not significant , at that stage of translation the program has been translated to a sequence of preprocessing tokens; the different amounts of whitespace in the output is just an artefact of how gcc -E works).






                      share|improve this answer





















                      • Thank you as well, but, unfortunately, I can only accept one answer, and 4386427 was too quick. :)
                        – jinscoe123
                        Nov 18 '18 at 8:39
















                      0














                      This example actually occurs in the Standard (C17 6.10/8):




                      EXAMPLE In: 



                      #define EMPTY
                      
EMPTY # include <file.h>


                      the sequence of preprocessing tokens on the second line is not a preprocessing directive, because it does not begin with a # at the start of translation phase 4, even though it will do so after the macro EMPTY has been replaced.




                      So the output you see from gcc -E is correct. (Note: the amount of whitespace here is not significant , at that stage of translation the program has been translated to a sequence of preprocessing tokens; the different amounts of whitespace in the output is just an artefact of how gcc -E works).






                      share|improve this answer





















                      • Thank you as well, but, unfortunately, I can only accept one answer, and 4386427 was too quick. :)
                        – jinscoe123
                        Nov 18 '18 at 8:39














                      0












                      0








                      0






                      This example actually occurs in the Standard (C17 6.10/8):




                      EXAMPLE In: 



                      #define EMPTY
                      
EMPTY # include <file.h>


                      the sequence of preprocessing tokens on the second line is not a preprocessing directive, because it does not begin with a # at the start of translation phase 4, even though it will do so after the macro EMPTY has been replaced.




                      So the output you see from gcc -E is correct. (Note: the amount of whitespace here is not significant , at that stage of translation the program has been translated to a sequence of preprocessing tokens; the different amounts of whitespace in the output is just an artefact of how gcc -E works).






                      share|improve this answer












                      This example actually occurs in the Standard (C17 6.10/8):




                      EXAMPLE In: 



                      #define EMPTY
                      
EMPTY # include <file.h>


                      the sequence of preprocessing tokens on the second line is not a preprocessing directive, because it does not begin with a # at the start of translation phase 4, even though it will do so after the macro EMPTY has been replaced.




                      So the output you see from gcc -E is correct. (Note: the amount of whitespace here is not significant , at that stage of translation the program has been translated to a sequence of preprocessing tokens; the different amounts of whitespace in the output is just an artefact of how gcc -E works).







                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered Nov 18 '18 at 8:32









                      M.MM.M

                      104k11114234




                      104k11114234












                      • Thank you as well, but, unfortunately, I can only accept one answer, and 4386427 was too quick. :)
                        – jinscoe123
                        Nov 18 '18 at 8:39


















                      • Thank you as well, but, unfortunately, I can only accept one answer, and 4386427 was too quick. :)
                        – jinscoe123
                        Nov 18 '18 at 8:39
















                      Thank you as well, but, unfortunately, I can only accept one answer, and 4386427 was too quick. :)
                      – jinscoe123
                      Nov 18 '18 at 8:39




                      Thank you as well, but, unfortunately, I can only accept one answer, and 4386427 was too quick. :)
                      – jinscoe123
                      Nov 18 '18 at 8:39


















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