Question

#define int_p int*
int_p p1,p2,p3; // only p1 is a pointer !

can somebody exlplain why it is so.

Answer 1

#define is just a textual substitution. The code above is equivalent to

int *p1, p2, p3;

so only p1 is a pointer. You need

typedef int* int_p;

instead.