c++: STL uses universal "end()" value for iterator? - c++

Does the ".end()" iterator for all instances of a given STL container point to the same "past-the-end" object?
e.g.
std::set<int> my_set_1;
// fill "my_set_1"
std::set<int> my_set_2;
// fill "my_set_2"
bool same_end_iterator = my_set_1.end() == my_set_2.end() ;
Is the last line implementation-dependent?

Does the ".end()" iterator for all instances of a given STL container point to the same "past-the-end" object?
No, the standard gives no guarantee of that, and in practice some container types probably won't work like that. For example std::vector::iterator is typically implemented as a pointer to an array element, with end() pointing past the end of the array.
Is the last line implementation-dependent?
The last line gives undefined behaviour. You can't compare iterators from different collections - including past-the-end iterators.

Conceptually, there's one off-the-end iterator per container and comparing iterators into different containers, even those of the same type, is UB.
In fact, for old school arrays, the off-the-end iterator is a+N where a is the base of the array and N is the number of elements in it, which means different off-the-end iterators for different arrays. This is also the most sensible implementation for std::vector::iterator that I can think of.

To my knowlegde, comparisons of iterators belonging to different containers is undefined, I don't recall which. This rather implies that given std::vector<int> foo, bar, foo.end() is not the same as bar.end().

Related

What is std::contiguous_iterator useful for?

For what purposes I can use it?
Why is it better than random_access_iterator?
Is there some advantage if I use it?
For a contiguous iterator you can get a pointer to the element the iterator is "pointing" to, and use it like a pointer to a contiguous array.
That can't be guaranteed with a random access iterator.
Remember that e.g. std::deque is a random-access container, but it's typically not a contiguous container (as opposed to std::vector which is both random access and contiguous).
In C++17, there is no such thing as a std::contiguous_iterator. There is the ContiguousIterator named requirement however. This represents a random access iterator over a sequence of elements where each element is stored contiguously, in exactly the same way as an array. Which means that it is possible, given a pointer to an value_type from an iterator, to perform pointer arithmetic on that pointer, which shall work in exactly the same way as performing the same arithmetic on the corresponding iterators.
The purpose of this is to allow for more efficient implementations of algorithms on iterators that are contiguous. Or to forbid algorithms from being used on iterators that aren't contiguous. One example of where this matters is if you're trying to pass C++ iterators into a C interface which is based on pointers to arrays. You can wrap such interfaces behind generic algorithms, verifying the contiguity of the iterator in the template.
Or at least, you could in theory; in C++17, that wasn't really possible.. The reason being that there was not actually a way to test if an iterator was a ContiguousIterator. There's no way to ask a pointer if doing pointer arithmetic on a pointer to an element from the iterator is legal. And there was no std::contiguous_iterator_category one could use for such iterators (as this could cause compatibility problems). So you couldn't use SFINAE tools to verify that an iterator was contiguous.
C++20's std::contiguous_iterator concept resolves this problem. It also resolves the other problem with contiguous iterators. See, the above explanation for ContiguousIterator's behavior starts with us having a pointer to an element from the range. Well, how did you get that? The obvious method would be to do something like std::addressof(*it), but what if it is the end iterator? The end iterator is not dereference-able, so you can't do that. Basically, even if you know that an iterator is contiguous, how do you go about converting it to the equivalent pointer?
The std::contiguous_iterator concept solves both of these problems. std::to_address is available, which will convert any contiguous iterator into its equivalent pointer value. And there is a traits tag that an iterator must provide to denote that it is in fact a contiguous iterator, just in case the default to_address implementation happens to be valid for a non-contiguous iterator.
A random access iterator only requires a constant time (iterator) + (offset), whereas contiguous iterators have the stronger guarantee that std::addressof(*((iterator) + (offset))) == std::addressof(*(iterator)) + (offset) (disregarding overloaded operator&s).
This basically means that the iterator is a pointer or a light wrapper around a pointer, so it is equivalent to a pointer to its elements, whereas a random access iterator can do more, at the cost of possibly being bulkier and being unable to turn it into a simple pointer.
As a C++20 Concept, I would expect you can use it to specify a different algorithm if the container is contiguous. Perhaps exploiting cache locality.

Are all pointers considered iterators?

This was a question on my exam and the answer is that all pointers are iterators but not all iterators are pointers. Why is this the case?
In a statement such as:
int *p = new int(4);
How can p be considered an iterator at all?
"Iterator" is some abstract concept, describing a certain set of operations a type must support with some specific semantics.
Pointers are iterators because they fulfill the concept iterator (and, even stronger, random access iterator), e.g. the operator++ to move to the next element and operator * to access the underlying element.
In your particular example, you get a standard iterator range with
[p, p+1)
which can be used for example in the standard algorithms, like any iterator pair. (It may not be particularly useful, but it is still valid.) The above holds true for all "valid" pointers, that is pointers that point to some object.
The converse implication however is false: For example, consider the std::list<T>::iterator. That is still an iterator, but it cannot be a pointer because it does not have an operator[].

Iterator equivalent to null pointer?

In an algorithm I'm currently implementing, I need to manipulate a std::list of struct T.
T holds a reference to another instance of T, but this reference can also be "unassigned".
At first, I wanted to use a pointer to hold this reference, but using an iterator instead makes it easier to remove from the list.
My question is : how to represent the equivalent to null pointer with my iterator?
I read general solution is to use myList.end(), but in my case, I need to test whether the iterator is "null" or not, and I may add or remove elements to the list between the moment when I store the iterator and the moment I remove it from list... Should I make the iterator point to a known list containing the "null" element? Or is there a more elegant solution?
According to this (emphasis by me):
Compared to the other base sequence
containers (vector and deque), lists
are the most efficient container doing
insertions at some position other than
the beginning or the end of the
sequence, and, unlike in these, all of
the previously obtained iterators and
references remain valid after the
insertion and refer to the same
elements they were referring before.
The same applies to erasure (with the obvious exception of iterators referring to a deleted element becoming invalidated). So yes, obtaining end() will always point to the same "invalid" element and should be safe to use.

Does a vector sort invalidate iterators?

std::vector<string> names;
std::vector<string>::iterator start = names.begin();
std::vector<string>::iterator end = names.end();
sort (start,end);
//are my start and end valid at this point?
//or they do not point to front and tail resp?
According to the C++ Standard §23.1/11:
Unless otherwise specified (either explicitly or by defining a function in terms of other functions), invoking
a container member function or passing a container as an argument to a library function shall not invalidate
iterators to, or change the values of, objects within that container.
§25.3 "Sorting and related operations" doesn't specify that iterators will be invalidated, so iterators in the question should stay valid.
They still point to the beginning and end. The values in those slots of the vector have probably changed, but the storage location in which each resides remains the same.
std::sort will not invalidate iterators to a vector. The sort template uses the * operator on the iterators to access and modify the contents of the vector, and modifying a vector element though an iterator to an element already in the vector will not invalidate any iterators.
In summary,
your existing iterators will not be invalidated
however, the elements they point to may have been modified
In addition to the support for the standard provided by Kirill V. Lyadvinsky (Does a vector sort invalidate iterators?):
25/5 "Algorithms library"
If an algorithm’s Effects section says
that a value pointed to by any
iterator passed as an argument is
modified, then that algorithm has an
additional type requirement: The type
of that argument shall satisfy the
requirements of a mutable iterator
(24.1).
24.1/4 "Iterator requirements"
Besides its category, a forward,
bidirectional, or random access
iterator can also be mutable or
constant depending on whether the
result of the expression *i behaves as
a reference or as a reference to a
constant.
std::vector keeps its elements in contiguous memory. std::sort takes arguments (iterators) by value and re-arranges the sequence between them. The net result is your local variables start and end are still pointing to first and one-past-the-last elements of the vector.

Is end() required to be constant in an STL map/set?

§23.1.2.8 in the standard states that insertion/deletion operations on a set/map will not invalidate any iterators to those objects (except iterators pointing to a deleted element).
Now, consider the following situation: you want to implement a graph with uniquely numbered nodes, where every node has a fixed number (let's say 4) of neighbors. Taking advantage of the above rule, you do it like this:
class Node {
private:
// iterators to neighboring nodes
std::map<int, Node>::iterator neighbors[4];
friend class Graph;
};
class Graph {
private:
std::map<int, Node> nodes;
};
(EDIT: Not literally like this due to the incompleteness of Node in line 4 (see responses/comments), but along these lines anyway)
This is good, because this way you can insert and delete nodes without invalidating the consistency of the structure (assuming you keep track of deletions and remove the deleted iterator from every node's array).
But let's say you also want to be able to store an "invalid" or "nonexistent" neighbor value. Not to worry, we can just use nodes.end()... or can we? Is there some sort of guarantee that nodes.end() at 8 AM will be the same as nodes.end() at 10 PM after a zillion insertions/deletions? That is, can I safely == compare an iterator received as a parameter to nodes.end() in some method of Graph?
And if not, would this work?
class Graph {
private:
std::map<int, Node> nodes;
std::map<int, Node>::iterator _INVALID;
public:
Graph() { _INVALID = nodes.end(); }
};
That is, can I store nodes.end() in a variable upon construction, and then use this variable whenever I want to set a neighbor to invalid state, or to compare it against a parameter in a method? Or is it possible that somewhere down the line a valid iterator pointing to an existing object will compare equal to _INVALID?
And if this doesn't work either, what can I do to leave room for an invalid neighbor value?
You write (emphasis by me):
§23.1.2.8 in the standard states that insertion/deletion operations on a set/map will not invalidate any iterators to those objects (except iterators pointing to a deleted element).
Actually, the text of 23.1.2/8 is a bit different (again, emphasis by me):
The insert members shall not affect the validity of iterators and references to the container, and the erase members shall invalidate only iterators and references to the erased elements.
I read this as: If you have a map, and somehow obtain an iterator into this map (again: it doesn't say to an object in the map), this iterator will stay valid despite insertion and removal of elements. Assuming std::map<K,V>::end() obtains an "iterator into the map", it should not be invalidated by insertion/removal.
This, of course, leaves the question whether "not invalidated" means it will always have the same value. My personal assumption is that this is not specified. However, in order for the "not invalidated" phrase to make sense, all results of std::map<K,V>::end() for the same map must always compare equal even in the face of insertions/removal:
my_map_t::iterator old_end = my_map.end();
// wildly change my_map
assert( old_end == my_map.end() );
My interpretation is that, if old_end remains "valid" throughout changes to the map (as the standard promisses), then that assertion should pass.
Disclaimer: I am not a native speaker and have a very hard time digesting that dreaded legaleze of the Holy PDF. In fact, in general I avoid it like the plague.
Oh, and my first thought also was: The question is interesting from an academic POV, but why doesn't he simply store keys instead of iterators?
23.1/7 says that end() returns an iterator that
is the past-the-end value for the container.
First, it confirms that what end() returns is the iterator. Second, it says that the iterator doesn't point to a particular element. Since deletion can only invalidate iterators that point somewhere (to the element being deleted), deletions can't invalidate end().
Well, there's nothing preventing particular collection implementation from having end() depend on the instance of collection and time of day, as long as comparisons and such work. Which means, that, perhaps, end() value may change, but old_end == end() comparison should still yield true. (edit: although after reading the comment from j_random_hacker, I doubt this paragraph itself evaluates to true ;-), not universally — see the discussion below )
I also doubt you can use std::map<int,Node>::iterator in the Node class due to the type being incomplete, yet (not sure, though).
Also, since your nodes are uniquely numbered, you can use int for keying them and reserve some value for invalid.
Iterators in (multi)sets and (multi)maps won't be invalidated in insertions and deletions and thus comparing .end() against previous stored values of .end() will always yield true.
Take as an example GNU libstdc++ implementation where .end() in maps returns the default intialized value of Rb_tree_node
From stl_tree.h:
_M_initialize()
{
this->_M_header._M_color = _S_red;
this->_M_header._M_parent = 0;
this->_M_header._M_left = &this->_M_header;
this->_M_header._M_right = &this->_M_header;
}
Assuming that (1) map implemented with red-black tree (2) you use same instance "after a zillion insertions/deletions"- answer "Yes".
Relative implmentation I can tell that all incarnation of stl I ever know use the tree algorithm.
A couple points:
1) end() references an element that is past the end of the container. It doesn't change when inserts or deletes change the container because it's not pointing to an element.
2) I think perhaps your idea of storing an array of 4 iterators in the Node could be changed to make the entire problem make more sense. What you want is to add a new iterator type to the Graph object that is capable of iterating over a single node's neighbours. The implementation of this iterator will need to access the members of the map, which possibly leads you down the path of making the Graph class extend the map collection. With the Graph class being an extended std::map, then the language changes, and you no longer need to store an invalid iterator, but instead simply need to write the algorithm to determine who is the 'next neighbour' in the map.
I think it is clear:
end() returns an iterator to the element one past the end.
Insertion/Deletion do not affect existing iterators so the returned values are always valid (unless you try to delete the element one past the end (but that would result in undefined behavior anyway)).
Thus any new iterator generated by end() (would be different but) when compared with the original using operator== would return true.
Also any intermediate values generated using the assignment operator= have a post condition that they compare equal with operator== and operator== is transitive for iterators.
So yes, it is valid to store the iterator returned by end() (but only because of the guarantees with associative containers, therefor it would not be valid for vector etc).
Remember the iterator is not necessarily a pointer. It can potentially be an object where the designer of the container has defined all the operations on the class.
I believe that this depends entirely on what type of iterator is being used.
In a vector, end() is the one past the end pointer and it will obviously change as elements are inserted and removed.
In another kind of container, the end() iterator might be a special value like NULL or a default constructed element. In this case it doesn't change because it doesn't point at anything. Instead of being a pointer-like thing, end() is just a value to compare against.
I believe that set and map iterators are the second kind, but I don't know of anything that requires them to be implemented in that way.
C++ Standard states that iterators should stay valid. And it is. Standard clearly states that in 23.1.2/8:
The insert members shall not affect the validity of iterators and references to the container, and the erase members shall invalidate only iterators and references to the erased elements.
And in 21.1/7:
end() returns an iterator which is the past-the-end value for the container.
So iterators old_end and new_end will be valid. That means that we could get --old_end (call it it1) and --new_end (call it it2) and it will be the-end value iterators (from definition of what end() returns), since iterator of an associative container is of the bidirectional iterator category (according to 23.1.2/6) and according to definition of --r operation (Table 75).
Now it1 should be equal it2 since it gives the-end value, which is only one (23.1.2/9). Then from 24.1.3 follows that: The condition that a == b implies ++a == ++b. And ++it1 and ++it2 will give old_end and new_end iterators (from definition of ++r operation Table 74). Now we get that old_end and new_end should be equal.
I had a similar question recently, but I was wondering if calling end() to retrieve an iterator for comparison purposes could possibly have race conditions.
According to the standard, two iterators are considered equivalent if both can be dereferenced and &*a == &*b or if neither can be dereferenced. Finding the bolded statement took a while and is very relevant here.
Because an std::map::iterator cannot be invalidated unless the element it points to has been removed, you're guaranteed that two iterators returned by end, regardless of what the state of the map was when they were obtained, will always compare to each other as true.