Scope of #define preprocessor in C

Question

The scope of #define is till the end of the file. But where does it start from. Basically I tried the following code.

 #include<stdio.h>  #include<stdlib.h>  #define pi 3.14  void fun();  int main() {  printf("%f \n",pi);  #define pi 3.141516     fun(); return 0; } void fun(){ printf("%f \n",pi);} 

The output of the above program comes out to be

3.140000 3.141416 

Considering preprocessing for main the value of pi should be 3.141516 and outside main 3.14. This is incorrect but please explain why.

Answer 1

The C preprocessor runs through the file top-to-bottom and treats #define statements like a glorified copy-and-paste operation. Once it encounters the line #define pi 3.14, it starts replacing every instance of the word pi with 3.14. The pre-processor does not process (or even notice) C-language scoping mechanisms like parenthesis and curly braces. Once it sees a #define, that definition is in effect until either the end of the file is reached, the macro is un-defined with #undef, or (as in this case) the macro is re-defined with another #define statement.

If you are wanting constants that obey the C scoping rules, I suggest using something more on the lines of const float pi = 3.14;.