Isn't the pattern supposed to be 2 2 3 3 ?

Question

Doubt regarding the inner working of pointers
https://code.sololearn.com/cevpej102B21/?ref=app

Answer 1

Step-by-step we have:

int array[] = {2,3,4,5};

int *arptr = array; // address of array[0]

int value= *arptr; // 2

cout<<value<<"\n"; // 2

value= *arptr++; // 2 (due to post-increment)

cout<<value<<"\n"; // 2

value= *arptr; // 3

cout<<value<<"\n"; // 3

value= *++arptr; // 4 (due to pre-increment)

cout<<value<<"\n"; // 4

Answer 2

Yes it does but `after` dereferencing and assigning the current element of the array to `value`, it'd be easy to illustrate it this way

value = *arptr;
++arptr;

Answer 3

If you tracking the program flow `before `

value= *++arptr;

you can see that the current address pointed by `arptr` is &array[0] + 1. When `++arptr` gets executed, first, the address gets incremented by 1, then gets dereferenced. So, now the `arptr == &array[0] + 2` and `*arptr == *(&array[0] + 2) == 4`. Actually, for the sake of simplicity, you could write it in two steps like so

++arptr;
value= *arptr;

Answer 4

Haha, glad to help out. 8D

Answer 5

*++arptr is nothing but
*(++arptr)
So, before the above operation the pointer gives the address of array[0] , after ++arptr , it should give the address of array[1] , and the value at i.e, *(++arptr) is the value of array[1] , which is 3 not 4.
I am maybe at a fundamental error at some point. Help me explain what is it.

Answer 6

Sorry for bothering you again..
Does *arptr++ influence the position of the pointer because that is only kind of updation that comes 'before'

Answer 7

Wow! Thanks. You guys at Sololearn are really prompt and to the point. Will surely rate it on Play. You saved my 4 marks from tomorrow's test.