What is the meaning of *(++p)?

Question

I wrote the following example:

#include <stdio.h>
int main()
{
    int p_arr[4]={3,5,6,1}, q_arr[4]={7,2,9,5};
    int *p=p_arr;
    int *q=q_arr;
    int i=1;
   printf("%d\t", *p + *q);  
   printf("%d\t", *(p++) + *(++q));   
   printf("%d\t", *(++p) + *(q++));
   printf("%d\t", *(p+i) + *q+i);
   printf("%d\t", *p+i + *(q+i));
   return 0;
}

Why is the fourth output 11 and the last output 12?

Answer 1

Pointers in C can be incremented or decremented. In the C semantics, incrementing a pointer means "make it point to the next element, as if the elements are in a array".

To make it even more clear, assume the following setup:

struct s_mystruct
{
    int a;
    double b;
};
struct s_mystruct v[5];
struct s_mystruct *p1 = v;
struct s_mystruct *p2 = v;
struct s_mystruct q1, q2;

Then it is true that *p1 == v[0];, i.e., dereferencing the pointer p1 is the same as accessing the first element of the vector v.

But since p1 is a vector, it can be incremented. It turns out that C has two pointer increment operators: pre-increment and post-increment. So, if:

p1 = v;
p2 = v;
q1 = *(p1++);
q2 = *(++p2);

then it is true that:

q1 == v[0];
q2 == v[1];
p1 == v + 1;
p2 == v + 1;

In other words, both statements increment the value of a pointer, and in the end, the pointer will be pointing to the next element, but p++ means "use the value that the pointer p had before incrementing it, and ++p means "use the value that the pointer p will have after it has been incremented.

Answer 2

Hopefully a little extra output will make things self-explanatory.

int main(void) {
  int p_arr[4] = {3, 5, 6, 1}, q_arr[4] = {7, 2, 9, 5};
  int *p = p_arr;
  int *q = q_arr;
  int i = 1;
  printf("\t");
  printf("i:%d p:%p,%d q:%p,%d\n", i, p, *p, q, *q);
  printf("%d\t", *p + *q);
  printf("i:%d p:%p,%d q:%p,%d\n", i, p, *p, q, *q);
  printf("%d\t", *(p++) + *(++q));
  printf("i:%d p:%p,%d q:%p,%d\n", i, p, *p, q, *q);
  printf("%d\t", *(++p) + *(q++));
  printf("i:%d p:%p,%d q:%p,%d\n", i, p, *p, q, *q);
  printf("%d\t", *(p + i) + *q + i);
  printf("i:%d p:%p,%d q:%p,%d\n", i, p, *p, q, *q);
  printf("%d\t", *p + i + *(q + i));
  printf("i:%d p:%p,%d q:%p,%d\n", i, p, *p, q, *q);
  return 0;
}

Output

    i:1 p:0xffffcaf0,3 q:0xffffcae0,7
10  i:1 p:0xffffcaf0,3 q:0xffffcae0,7
5   i:1 p:0xffffcaf4,5 q:0xffffcae4,2
8   i:1 p:0xffffcaf8,6 q:0xffffcae8,9
11  i:1 p:0xffffcaf8,6 q:0xffffcae8,9
12  i:1 p:0xffffcaf8,6 q:0xffffcae8,9