Why is the end() element in Associative container print the same value as the last element?

Question

In the code below, the value that the iterator points to is the same for both the last and the second last element.

#include <iostream>
#include <set>
using namespace std;


int main() 
{
    set<int> s1 = {4,3,2,5,1};
    set<int>::iterator i;
    
    i = s1.end();
    cout << *i << endl; // 5
    
    i--;
    cout << *i << endl; // 5
    
    
    cout << *s1.end() << endl;  // 5
    cout << *(--s1.end()) << endl;  // 5
    
    return 0;
}

In my understanding the value pointed to by the end element should be null. Why is this so?

*"In my understanding the value pointed to by the end element should be null"* - even if that were true (which it is not), dereferencing a null pointer invokes *undefined behavior* — UnholySheep, Dec 10 '20 at 13:55
The target is for `vector`, but it's the same thing really, there's nothing special about associative containers. — cigien, Dec 10 '20 at 14:05

Tony Tannous · Accepted Answer · 2020-12-10T13:58:32.283

6

You invoked undefined behavior, std::set::end

Returns an iterator to the element following the last element of the set. This element acts as a placeholder; attempting to access it results in undefined behavior.

Undefined behavior renders the entire program meaningless.

edited Dec 10 '20 at 13:58

answered Dec 10 '20 at 13:53

Tony Tannous

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463035818_is_not_an_ai · Answer 2 · 2020-12-10T14:09:41.817

In my understanding the value pointed to by the end element should be null. Why is this so?

Why your understanding is wrong I cannot tell ;). Snarkyness aside: No. There is no "end element". The end iterator points to one past the last element. You cannot dereference it. If you do you invoke undefined behavior.

The last element in a container is typically refered to as the "back" and many containers have a back() method to access it.

score 0 · Answer 3 · answered Dec 10 '20 at 14:01

In my understanding the value pointed to by the end element should be null. Why is this so?

It isn't.

The one-past-the-end iterator doesn't "point" to anything; it's not dereferencable.

C-strings might appear at first to be an exception to this rule, but they're not: they are logically terminated by a NULL character that you can check for, but that NULL is still part of an array of char objects, and you cannot dereference a char* that's the one-past-the-end iterator for that array. Think of a C-string's null terminator as a "this page deliberately left blank"; the page doesn't give you any more story, but it's still a page.

Why is the end() element in Associative container print the same value as the last element?

That's one possible result of a program with undefined behaviour.

Why is the end() element in Associative container print the same value as the last element?

3 Answers3