#define constants

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  • Ioannis Vranos
    #1

    #define constants

    I think I read somewhere that #define constants in the style


    #define SOMETHING 6


    are evaluated to long or unsigned long, depending on the situation and
    not int.
    Tags: None
    • Ioannis Vranos
      #2
      Re: #define constants

      Ioannis Vranos wrote:
      I think I read somewhere that #define constants in the style
      >
      >
      #define SOMETHING 6
      >
      >
      are evaluated to long or unsigned long, depending on the situation and
      not int.

      .... Is that true?

        Comment

        • Victor Bazarov
          #3
          Re: #define constants

          Ioannis Vranos wrote:
          Ioannis Vranos wrote:
          >I think I read somewhere that #define constants in the style
          >>
          >>
          >#define SOMETHING 6
          >>
          >>
          >are evaluated to long or unsigned long, depending on the situation
          >and not int.
          >
          >
          ... Is that true?
          I don't know whether it's true that you read that nonsense somewhere
          or whether you dreamed it. I know that an integral literal has the
          type 'int' if it fits in it, and if it doesn't, it has type long int,
          unless overridden by a suffix (absent in your example). This all is
          defined in [lex.icon]/2.

          V
          --
          Please remove capital 'A's when replying by e-mail
          I do not respond to top-posted replies, please don't ask


            Comment

            • Ioannis Vranos
              #4
              Re: #define constants

              Victor Bazarov wrote:
              Ioannis Vranos wrote:
              >Ioannis Vranos wrote:
              >>I think I read somewhere that #define constants in the style
              >>>
              >>>
              >>#define SOMETHING 6
              >>>
              >>>
              >>are evaluated to long or unsigned long, depending on the situation
              >>and not int.
              >>
              >... Is that true?
              >
              I don't know whether it's true that you read that nonsense somewhere
              or whether you dreamed it. I know that an integral literal has the
              type 'int' if it fits in it, and if it doesn't, it has type long int,
              unless overridden by a suffix (absent in your example). This all is
              defined in [lex.icon]/2.

              My mistake. It is about the "identifier " of the macro expressions

              defined identifier

              and

              defined ( identifier)


              Errata for K&R2 page 232 (from
              http://www-db-out.research.bell-labs.com/cm/cs/cbook/2ediffs.html):\



              232(§A12.5): The result of the defined operator is not replaced
              literally by 0L or 1L, nor are undefined names literally by 0L, but just
              by plain 0 or 1. However, the constant expression is nevertheless
              evaluated as if these and other constants appearing have long or
              unsigned long type.

                Comment

                • Ioannis Vranos
                  #5
                  Re: #define constants

                  Ioannis Vranos wrote:
                  Victor Bazarov wrote:
                  >Ioannis Vranos wrote:
                  >>Ioannis Vranos wrote:
                  >>>I think I read somewhere that #define constants in the style
                  >>>>
                  >>>>
                  >>>#define SOMETHING 6
                  >>>>
                  >>>>
                  >>>are evaluated to long or unsigned long, depending on the situation
                  >>>and not int.
                  >>... Is that true?
                  >I don't know whether it's true that you read that nonsense somewhere
                  >or whether you dreamed it. I know that an integral literal has the
                  >type 'int' if it fits in it, and if it doesn't, it has type long int,
                  >unless overridden by a suffix (absent in your example). This all is
                  >defined in [lex.icon]/2.
                  >
                  >
                  My mistake. It is about the "identifier " of the macro expressions
                  >
                  defined identifier
                  >
                  and
                  >
                  defined ( identifier)
                  >
                  >
                  Errata for K&R2 page 232 (from
                  http://www-db-out.research.bell-labs.com/cm/cs/cbook/2ediffs.html):\

                  >
                  >
                  232(§A12.5): The result of the defined operator is not replaced
                  literally by 0L or 1L, nor are undefined names literally by 0L, but just
                  by plain 0 or 1. However, the constant expression is nevertheless
                  evaluated as if these and other constants appearing have long or
                  unsigned long type.

                  The fact that the "identifier " in a "defined" preprocessor statement is
                  evaluated as long or unsigned long, doesn't it imply that this
                  identifier which was defined with #define" preprocessor statement is
                  evaluated as long or unsigned long everywhere?

                    Comment

                    • Ioannis Vranos
                      #6
                      Re: #define constants

                      Ioannis Vranos wrote:
                      Ioannis Vranos wrote:
                      >Victor Bazarov wrote:
                      >>Ioannis Vranos wrote:
                      >>>Ioannis Vranos wrote:
                      >>>>I think I read somewhere that #define constants in the style
                      >>>>>
                      >>>>>
                      >>>>#define SOMETHING 6
                      >>>>>
                      >>>>>
                      >>>>are evaluated to long or unsigned long, depending on the situation
                      >>>>and not int.
                      >>>... Is that true?
                      >>I don't know whether it's true that you read that nonsense somewhere
                      >>or whether you dreamed it. I know that an integral literal has the
                      >>type 'int' if it fits in it, and if it doesn't, it has type long int,
                      >>unless overridden by a suffix (absent in your example). This all is
                      >>defined in [lex.icon]/2.
                      >>
                      >My mistake. It is about the "identifier " of the macro expressions
                      >>
                      >defined identifier
                      >>
                      >and
                      >>
                      >defined ( identifier)
                      >>
                      >>
                      >Errata for K&R2 page 232 (from
                      >http://www-db-out.research.bell-labs...ediffs.html):\
                      >>
                      >>
                      >232(§A12.5): The result of the defined operator is not replaced
                      >literally by 0L or 1L, nor are undefined names literally by 0L, but just
                      >by plain 0 or 1. However, the constant expression is nevertheless
                      >evaluated as if these and other constants appearing have long or
                      >unsigned long type.
                      >
                      >
                      The fact that the "identifier " in a "defined" preprocessor statement is
                      evaluated as long or unsigned long, doesn't it imply that this
                      identifier which was defined with #define" preprocessor statement is
                      evaluated as long or unsigned long everywhere?

                      My mistake again, the _result_ of the defined operator is replaced by 0
                      or 1 which are evaluated as 0L and 1L. Strange stuff.

                        Comment

                        • Ioannis Vranos
                          #7
                          Re: #define constants

                          OK, lets get it from the start again:


                          Errata for K&R2 page 232 (from
                          http://www-db-out.research.bell-labs.com/cm/cs/cbook/2ediffs.html):\



                          232(§A12.5): The result of the defined operator is not replaced
                          literally by 0L or 1L, nor are undefined names literally by 0L, but just
                          by plain 0 or 1. However, the constant expression is nevertheless
                          evaluated as if these and *other constants* appearing have long or
                          unsigned long type.


                          When it mentions "these and other constants", what are the other
                          constants implied?

                            Comment

                            • Ron Natalie
                              #8
                              Re: #define constants

                              Ioannis Vranos wrote:
                              I think I read somewhere that #define constants in the style
                              >
                              >
                              #define SOMETHING 6
                              >
                              >
                              are evaluated to long or unsigned long, depending on the situation and
                              not int.
                              Nope. Not true. #define has little to do with (unless you are
                              throwing in some token pasting operators), an decimal integer literal
                              has type the smaller of int or long int that can hold it.

                              Other constraints on the language make 6 HAVE to be of type int.

                              There's no way a decimal literal ever has type unsigned.

                                Comment

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