I was revisiting pointers when I had this doubt.
int *ptr;
int arr[5] = {10,20,30,40,50};
ptr = &arr[0];
Now printf(\"Value: %d\"
The first snippet is obvious: it prints what ptr
points to, i.e. 10.
The second one, moves the pointer forward of one element, which then points to the next element, i.e. 20.
The third snippet is exactly the same as the previous one, because its first instruction increments the pointer and returns the unincremented value, which is dereferenced, but its result is discarded; what is dereferenced in the printf
is the incremented pointer, which now points to 30.
The last snippet is different: ++*ptr
is ++(*ptr)
; *ptr
dereferences ptr
(which already points to 30
), yielding 30, and ++
increments such value, which becomes 31.