Incrementing pointer (ptr++) and (*ptr++)

Question

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\"

Answer 1

The trick to this is that whenever both are unary operators, then associativity is from right to left.

*ptr++ --> first increment pointer to next location and then dereference the value at that location.

Similarly *++ptr -->first increment pointer and then dereference.

++*ptr --> first dereference value at ptr and then increment the value.

Answer 2

Because of precedence.

• *ptr++ is the same as *(ptr++), the pointer is incremented, but its previous value is dereferenced.
• ptr++ is the same as, well, ptr++.
• ++*ptr is the same as ++(*ptr), i.e. it increments the thing being pointed to, not the pointer.

Answer 3

The order of C Precedence from high priority:

()
++
* & for pointers
* /
+ -

so,

 *ptr++ equivlant to *(ptr++)
 ++*ptr equivlant to ++(*ptr)

and for this line *ptr++; it will only increment the ptr pointer.

Answer 4

When you do * ptr++, you are actually doing * (ptr++), so the pointer get incremented, not the variable it points to

When ++ *ptr, it's actually ++(* ptr), so it increments the value returned by the pointer

Answer 5

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.