How p4 is giving output 1 in this code can anyone explain this to me ?

Question

#include<stdio.h>
#define cp_d char*
typedef char* cp_t;
int main(){
cp_t p1,p2;
cp_d p3,p4;
printf("%d %d\n",sizeof(p1),sizeof(p2));
printf("%d %d\n",sizeof(p3),sizeof(p4));
return 0;
}

Answer 1

Because `cp_d` in main function is replaced by `char*` before the compilation.

The line 👇

cp_d p3, p4;

Will be replaced by 👇before compilation

char* p3, p4;

which means, <p3> is a pointer, but <p4> is plain char (not a pointer).

In case of `cp_t`, both <p1> and <p2> are both defined as `char*` because `typedef` defined a type alias.