C - SizeOf Pointers

Question

char c[] = {\'a\',\'b\',\'c\'};
int* p = &c[0];
printf(\"%i\\n\", sizeof(*p)); //Prints out 4
printf(\"%i\\n\", sizeof(*c)); //Prints out 1

I a

Answer 1

sizeof(*p) will print size of p which is 4 because of int but c is of char that's why it is 1