Scope of macros in C?

Question

How these macros are evaluated?

# define i 20 void fun();  int main(){   printf("%d",i);   fun();   printf("%d",i);   return 0; }  void fun(){   #undef i   #define i 30 } 

gives output as 2020 but whereas,

# define i 20 void fun(){   #undef i   #define i 30 }  int main(){   printf("%d",i);   fun();   printf("%d",i);   return 0; } 

gives output as 3030. Please Explain. Thanks.

Answer 1

C Preprocessor works top to bottom irrespective of function calls. It is effective from that point (line) in whatever the file that macro is defined, until corresponding undef or till end of the translation unit.

So, your code would become,

# define i 20                // from now on, all token i should become 20 void fun(); int main() {   printf("%d",i);   // printf("%d",20);   fun();   printf("%d",i);   // printf("%d",20);   return 0; } void fun() { #undef i               // from now on, forget token i #define i 30               // from now on, all token i should become 30 } 

Your second code would become,

# define i 20                // from now on, all token i should become 20 void fun() { #undef i                // from now on, forget i #define i 30                // from now on, all token i should become 30 } int main() {   printf("%d",i);    //  printf("%d",30);   fun();   printf("%d",i);    // printf("%d",30);   return 0; } 

Answer 2

There's no scope involved at all. Macros are handled at the preprocessing stage separately and independently from the compilation stage and have no concept of a C scope. Your examples could just as easily be:

#define i 20  void fun();  int main() {   printf("%d",i);   fun();   printf("%d",i);   return 0; }  void fun() { }  #undef i #define i 30 

And:

#define i 20 #undef i #define i 30  void fun() { }  int main() {   printf("%d",i);   fun();   printf("%d",i);   return 0; } 

You can see from those why it behaves the way it does.