Preprocessor part 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
#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