While incrementing a pointer which points to an array, why does my array feels be reversed?

Question

When I'm dereferencing and printing the output of a given array pointer, I'm getting my array in reverse. Here's the code

using namespace std;

int main()
{   
    int arr[5] = {1,2,3,4};
    int* ptr = arr;
    cout<<ptr<<endl;
    cout<<*ptr++<<endl;
    cout<<*ptr<<" "<<*ptr++<<" "<<*ptr++<<" "<<*ptr++;
    return 0;
}

Output

0x61fee8
1
0 4 3 2

But since I filled the array in increasing order and incremented the pointer I was expecting the output to be as 1 2 3 4

and why there is a zero in 2nd line?

Does this answer your question? https://stackoverflow.com/questions/4176328/undefined-behavior-and-sequence-points — 463035818_is_not_an_ai, Mar 08 '22 at 12:56
TL;DR Configure your compiler to use C++17, or split the last `cout` to separate statements. — HolyBlackCat, Mar 08 '22 at 12:57
Does this answer your question? [Undefined behavior and sequence points](https://stackoverflow.com/questions/4176328/undefined-behavior-and-sequence-points) — prapin, Mar 08 '22 at 13:16

score 1 · Accepted Answer · edited Mar 12 '22 at 21:25

1

The first thing to note is that the size of the array is 5 and the number of initialization elements is 4, which means that the last element will be initialized to 0.

The reason for unordered output is that C++ compiler does not guarantee the order in which the parameters of << are evaluated. It means that, in your last cout, it is possible that the third ptr++ is evaluated before the first one.

If you replace your last cout with multiple couts as follows, you will see that the code works as you expect.

cout<< *ptr;
cout<< *ptr++;
cout<< *ptr++;
cout<< *ptr++;

edited Mar 12 '22 at 21:25

halfer

19,824
17
99
186

answered Mar 08 '22 at 13:13

moriaz

96
6

1

which order would result in `0` being the first number to be printed ? – 463035818_is_not_an_ai Mar 08 '22 at 13:19
@463035818_is_not_a_number, For example, the reverse order. when the *ptr is evaluated after all "*ptr++"s are evaluated. In that case, the fifth element of the array (which does not exist) is printed, and since the value that is stored in that location is unknown, it is probably 0 in the specific run that the questioner has sent. Note that the code in total has four "ptr++"s. – moriaz Mar 08 '22 at 13:27
3

You should mention that this was fixed in C++17. – HolyBlackCat Mar 08 '22 at 13:37
yeah, sorry for not being clear. I actually wanted to point out that the answer is incomplete. 4 increments and dereferencing the pointer is undefined behavior – 463035818_is_not_an_ai Mar 08 '22 at 13:44
@HolyBlackCat: I wouldn't say "fixed" but rather specified, mandated, or defined. – Bathsheba Mar 08 '22 at 13:53
@463035818_is_not_a_number: You sure about that: `arr[4]` is zero. – Bathsheba Mar 08 '22 at 13:53
@Bathsheba oh I am not so sure anymore – 463035818_is_not_an_ai Mar 08 '22 at 14:21
@HolyBlackCat - what was fixed and in what way? – pm100 Mar 12 '22 at 21:29
@pm100 See rule 19 [here](https://en.cppreference.com/w/cpp/language/eval_order). In short, the first operand of `<<` and `>>` is now sequenced before the second operand. – HolyBlackCat Mar 12 '22 at 21:35
@HolyBlackCat ty – pm100 Mar 12 '22 at 21:38
I got it, Thanks for your answers and inpouts fellas – Alpha Sep 29 '22 at 11:06

While incrementing a pointer which points to an array, why does my array feels be reversed?

1 Answers1