0
Can someone explain how this code works?
Hi all, in a C-challange i stumpled across this question: // begin of code #include<stdio.h> // macro definitions: #define macro(r,a,b) b##a##r #define CHAOS macro(r,a,b) // function forward declaritions: void CHAOS(int n); // main function int main(void) { bar(4); bar(2); return 0; } // function definitions: void CHAOS(int n) { printf("%d", n); } // end of code 42 is the output. I debugged the code, because i did not understand how it worked and the function CHAOS() is called at bar(). Why is that?
1 Resposta
+ 3
#define is a pre-processor directive first CHOAS is replaced by macro(r,a,b)
Again we have another #define it replaces macro(r,a,b) by b##a##r
## acts as token pasting operator
So b##a##r=bar
Hence void CHOAS(int n) is void bar(int n)
bar(4);//outputs 4
bar(2);//outputs 2
So,final output is 42