C Preprocessor-3

Operators in Preprocessor:

• The 2 operators used in the preprocessor are

1. #
2. ##

1.# operator:

• If the macros are present with in the quoted strings it is not replaced by the #defined macro.
• But if the macroname is preceded by a # then the macro is expanded into a quoted string with the parameter replaced by the actual argument
• Eg:
  #include<stdio.h>
  #define toprint(expr) printf(#expr "=%d\n",expr)
  int main()
  {
      int x=10,y=5;
      toprint(x/y);
      return 0;
  }
  when toprint(x/y);is invoked the macro is expanded in to
      printf("x/y" "=%d\n",x/y);
  and the strings are concatenated.so the statement becomes
       printf("x/y=%d\n",x/y);
  The output of the above program is x/y=2
• In the above example instead of #expr if only expr is present like
      #define toprint(expr) printf("expr=%d\n",expr) then the output is expr=2

2.## operator:

• The preprocessor operator ## provides a way to concatenate actual arguments during macro expansion.
• If a parameter in the repacement text is adjacent to a ##, the parameter is replaced by the actual argument,the ## and surrounding white space are removed,and the result is rescanned
• Example The macro Join concatenates its two arguments
      #define Join(front,back) front##back
  now Join(hello,100) resuts hello100.

NULL Directive

• # 
• A preprocessor line of the form # has no effect 
• if the line has only #symbol the preprocessor ignores the line