How the address of a variable is affected by arithmetic oprators

Question

c = 500; //let address of c be 2004
ptr = &c;
printf("%d", ptr + 1);
printf("%d ", *(ptr + 2));
printf("%d", ptr - 1); // so what will be outputs and why??

Answer 1

Depends on the type of c. Assuming it is an int and is 4 bytes, the first output will be 2008. The second will be undefined behaviour because you are reading at an invalid memory address. The third will print 2000. This is because adding 1 to pre will increase its value by sizeof int which is 4

Answer 2

@The Man thanks for answering. So, according to you the first one will be 2008??