Are the end iterators always in the rhs of the comparison? - c++

From what I have found in my testing range-based standard algorithms and containers always do the following comparison it!=c.end() aka the end iterator is on the right of the != operator and I want to know if there's an standard or some warranty that the following will not appear on the STL implementation c.end()!=it
I'm curious because I'm building an iterators library and my != operator looks something like this
bool operator!=(const Iterator& lhs, const Iterator& rhs)
{
return lhs._validate();
}
Which is valid as long as the rhs is always the end iterator

When ever it!=c.end() is false, technically end iterator is on the both sides of the comparison.
There is no requirement to write it!=c.end() instead of c.end()!=it nor is there a guarantee that latter won't be written. Both are equally allowed. Furthermore it is entirely possible that neither iterator of the comparison is an end iterator: c.begin()!=c.begin() is perfectly fine. As such, your suggestion won't work.
Writing the changing "variable" on the left and the "constant" with a known property on the right is conventional in programming as well as natural in relation to English: We ask "Is the iterator at the end?" rather than "Is at the end the iterator?".

23.3.1 In general [iterator.requirements.general]
1 Iterators are a generalization of pointers that allow a C ++ program to work with different data structures (for example, containers and ranges) in a uniform manner.
2 Since iterators are an abstraction of pointers, their semantics are a generalization of most of the semantics of pointers in C ++ . This ensures that every function template that takes iterators works as well with regular pointers.
Comparison of pointers is well-defined even if any of them is a null pointer or a past-the-end pointer. They have commutative property, that is result stays the same when the order of arguments is reversed. So comparison functions working with iterators are expected to behave in similar manner and not just randomly dereference arguments or require one of them not to be past-the-end iterator.

Related

Are end+1 iterators for std::string allowed?

Is it valid to create an iterator to end(str)+1 for std::string?
And if it isn't, why isn't it?
This question is restricted to C++11 and later, because while pre-C++11 the data was already stored in a continuous block in any but rare POC toy-implementations, the data didn't have to be stored that way.
And I think that might make all the difference.
The significant difference between std::string and any other standard container I speculate on is that it always contains one element more than its size, the zero-terminator, to fulfill the requirements of .c_str().
21.4.7.1 basic_string accessors [string.accessors]
const charT* c_str() const noexcept;
const charT* data() const noexcept;
1 Returns: A pointer p such that p + i == &operator[](i) for each i in [0,size()].
2 Complexity: Constant time.
3 Requires: The program shall not alter any of the values stored in the character array.
Still, even though it should imho guarantee that said expression is valid, for consistency and interoperability with zero-terminated strings if nothing else, the only paragraph I found casts doubt on that:
21.4.1 basic_string general requirements [string.require]
4 The char-like objects in a basic_string object shall be stored contiguously. That is, for any basic_string object s, the identity &*(s.begin() + n) == &*s.begin() + n shall hold for all values of n such that 0 <= n < s.size().
(All quotes are from C++14 final draft (n3936).)
Related: Legal to overwrite std::string's null terminator?
TL;DR: s.end() + 1 is undefined behavior.
std::string is a strange beast, mainly for historical reasons:
It attempts to bring C compatibility, where it is known that an additional \0 character exists beyond the length reported by strlen.
It was designed with an index-based interface.
As an after thought, when merged in the Standard library with the rest of the STL code, an iterator-based interface was added.
This led std::string, in C++03, to number 103 member functions, and since then a few were added.
Therefore, discrepancies between the different methods should be expected.
Already in the index-based interface discrepancies appear:
§21.4.5 [string.access]
const_reference operator[](size_type pos) const;
reference operator[](size_type pos);
1/ Requires: pos <= size()
const_reference at(size_type pos) const;
reference at(size_type pos);
5/ Throws: out_of_range if pos >= size()
Yes, you read this right, s[s.size()] returns a reference to a NUL character while s.at(s.size()) throws an out_of_range exception. If anyone tells you to replace all uses of operator[] by at because they are safer, beware the string trap...
So, what about iterators?
§21.4.3 [string.iterators]
iterator end() noexcept;
const_iterator end() const noexcept;
const_iterator cend() const noexcept;
2/ Returns: An iterator which is the past-the-end value.
Wonderfully bland.
So we have to refer to other paragraphs. A pointer is offered by
§21.4 [basic.string]
3/ The iterators supported by basic_string are random access iterators (24.2.7).
while §17.6 [requirements] seems devoid of anything related. Thus, strings iterators are just plain old iterators (you can probably sense where this is going... but since we came this far let's go all the way).
This leads us to:
24.2.1 [iterator.requirements.general]
5/ Just as a regular pointer to an array guarantees that there is a pointer value pointing past the last element of the array, so for any iterator type there is an iterator value that points past the last element of a corresponding sequence. These values are called past-the-end values. Values of an iterator i for which the expression *i is defined are called dereferenceable. The library never assumes that past-the-end values are dereferenceable. [...]
So, *s.end() is ill-formed.
24.2.3 [input.iterators]
2/ Table 107 -- Input iterator requirements (in addition to Iterator)
List for pre-condition to ++r and r++ that r be dereferencable.
Neither the Forward iterators, Bidirectional iterators nor Random iterator lift this restriction (and all indicate they inherit the restrictions of their predecessor).
Also, for completeness, in 24.2.7 [random.access.iterators], Table 111 -- Random access iterator requirements (in addition to bidirectional iterator) lists the following operational semantics:
r += n is equivalent to [inc|dec]rememting r n times
a + n and n + a are equivalent to copying a and then applying += n to the copy
and similarly for -= n and - n.
Thus s.end() + 1 is undefined behavior.
Returns: A pointer p such that p + i == &operator[](i) for each i in [0,size()].
std::string::operator[](size_type i) is specified to return "a reference to an object of type charT with value charT() when i == size(), so we know that that pointer points to an object.
5.7 states that "For the purposes of [operators + and -], a pointer to a nonarray object behaves the same as a pointer to the first element of an array of length one with the type of the object as its element type."
So we have a non-array object and the spec guarantees that a pointer one past it will be representable. So we know std::addressof(*end(str)) + 1 has to be representable.
However that's not a guarantee on std::string::iterator, and there is no such guarantee anywhere in the spec, which makes it undefined behavior.
(Note that this is not the same as 'ill-formed'. *end(str) + 1 is in fact well-formed.)
Iterators can and do implement checking logic that produce various errors when you do things like increment the end() iterator. This is in fact what Visual Studios debug iterators do with end(str) + 1.
#define _ITERATOR_DEBUG_LEVEL 2
#include <string>
#include <iterator>
int main() {
std::string s = "ssssssss";
auto x = std::end(s) + 1; // produces debug dialog, aborts program if skipped
}
And if it isn't, why isn't it?
for consistency and interoperability with zero-terminated strings if nothing else
C++ specifies some specific things for compatibility with C, but such backwards compatibility is limited to supporting things that can actually be written in C. C++ doesn't necessarily try to take C's semantics and make new constructs behave in some analogous way. Should std::vector decay to an iterator just to be consistent with C's array decay behavior?
I'd say end(std) + 1 is left as undefined behavior because there's no value in trying to constrain std::string iterators this way. There's no legacy C code that does this that C++ needs to be compatible with and new code should be prevented from doing it.
New code should be prevented from relying on it... why? [...] What does not allowing it buy you in theory, and how does that look in practice?
Not allowing it means implementations don't have to support the added complexity, complexity which provides zero demonstrated value.
In fact it seems to me that supporting end(str) + 1 has negative value since code that tries to use it will essentially be creating the same problem as C code which can't figure out when to account for the null terminator or not. C has enough off by one buffer size errors for both languages.
A std::basic_string<???> is a container over its elements. Its elements do not include the trailing null that is implicitly added (it can include embedded nulls).
This makes lots of sense -- "for each character in this string" probably shouldn't return the trailing '\0', as that is really an implementation detail for compatibility with C style APIs.
The iterator rules for containers were based off of containers that don't shove an extra element at the end. Modifying them for std::basic_string<???> without motivation is questionable; one should only break a working pattern if there is a payoff.
There is every reason to think that pointers to .data() and .data() + .size() + 1 are allowed (I could imagine a twisted interpretation of the standard that would make it not allowed). So if you really need read-only iterators into the contents of a std::string, you can use pointer-to-const-elements (which are, after all, a kind of iterator).
If you want editable ones, then no, there is no way to get a valid iterator to one-past-the-end. Neither can you get a non-const reference to the trailing null legally. In fact, such access is clearly a bad idea; if you change the value of that element, you break the std::basic_string's invariant null-termination.
For there to be an iterator to one-past-the-end, the const and non-const iterators to the container would have to have a different valid range, or a non-const iterator to the last element that can be dereferenced but not written to must exist.
I shudder at making such standard wording watertight.
std::basic_string is already a mess. Making it even stranger would lead to standard bugs and would have a non-trivial cost. The benefit is really low; in the few cases where you want access to said trailing null in an iterator range, you can use .data() and use the resulting pointers as iterators.
I can't find a definitive answer, but indirect evidence points at end()+1 being undefined.
[string.insert]/15
constexpr iterator insert(const_iterator p, charT c);
Preconditions: p is a valid iterator on *this.
It would be unreasonable to expect this to work with end()+1 as the iterator, and it indeed causes a crash on both libstdc++ and libc++.
This means end()+1 is not a valid iterator, meaning end() is not incrementable.

Example of Input Iterator where `end()` actually denotes one-past-the end?

I'm currently trying to make sense of some ideas wrt. C++ iterators, and I have been wondering ...
Given an Incremental / Single Pass / Input / Output Iterator, can there actually exist such a thing as a one-past-the-end position/element for such an Iterator, or are all InputIterator end() Iterators "naturally" some form of singular values that are treated specially by operator==?
I think what I mean is this: For anything from ForwardIterator on "upwards", it can make total sense to have a trivial operator== that just checks whether the two iterator objects, regardless of end-ness, point to the same element. Can this ever make sense for an InputIterator?
An InputIterator that is not a ForwardIterator is one for which incrementing it invalidates the prior value (meaning any iterator with the same value, i.e. any copies of the original).
In general it is only valid to compare iterators from "the same sequence" (that is, were one is reachable from the other). For the iterators you're talking about, this means the only valid comparisons are between:
two equal non-end iterators
two end iterators
an end iterator and a non-end iterator
You can't (by the guarantees of this interface) compare two unequal non-end iterators because you never have two valid non-end iterators where one is reachable from the other. The one that's "behind" has already been invalidated.
So it seems likely that you could implement the iterator such that it contains a data member that has one value in end iterators and a different value in non-end iterators. For the typical example of a stream iterator, that data member could be bool isEndOfStream. Then operator== would not need to contain any special case code, it just needs to compare that field. It would then be natural for all end iterators to be interchangeable: that field is the only thing on them that will ever be used.
It might well be efficient, since iterators become end iterators far less frequently than iterators are compared, so a write in the rare case to allow the common case to be just a read and compare would seem sensible. Such an iterator comparison would return true for any two non-end iterators, but that's fine because either they're genuinely equal (in which case returning true is correct) or else comparing them is not valid (in which case behavior is undefined).
The canonical example is the istream_iterator (template), which becomes singular when the underlying stream extraction fails. This can be detected by comparing against a default-constructed iterator of the same type, which is equivalent to a singular iterator. For example:
std::vector<int> v(std::istream_iterator<int>(std::cin), {});
This is the equivalent of:
std::vector<int> v;
for (int n; std::cin >> n; ) { v.push_back(n); }
To stress again: all one-past-the-end istream iterators are equivalent, independent of the stream they came from. These iterators are an example where "singular" (= not associated with any container) and "one-past-the-end" (= result of incrementing the last dereferenceable iterator) mean the same thing.

Decrementing an off the end iterator

I was reading today about how for containers that support bidirectional iteration, this piece of code is valid:
Collection c(10, 10);
auto last = --c.end();
*last;
That got me thinking, is it required that when submitting a pair of bidirectional iterators [beg, end) to an algorithm in the STL that --end is defined? If so, should the result be dereferenceable?
ie
void algo(T beg, T end){
//...
auto iter = --end;
//...
*iter;
}
If the algorithm requires a range defined by bidirectional iterators first and last, then --last needs to be valid under the same conditions that ++first does -- namely that the range isn't empty. The range is empty if and only if first == last.
If the range isn't empty, then --last evaluates to an iterator that refers to the last element in the range, so *--last indeed also needs to be valid.
That said, there aren't all that many standard algorithms that require specifically a bidirectional iterator (and don't require random-access). prev, copy_backward, move_backward, reverse, reverse_copy, stable_partition, inplace_merge, [prev|next]_permutation.
If you look at what some of those do, you should see that typically the algorithm does decrement the end-of-range iterator and dereference the result.
As James says, for containers the function end() returns an iterator by value. There is no general requirement that for iterators that --x should be a well-formed expression when x is an rvalue of the type. For example, pointers are bidirectional iterators, and a function declared as int *foo(); returns a pointer by value, and --foo() is not a well-formed expression. It just so happens that for the containers you've looked at in your implementation, end() returns a class type which has operator-- defined as a member function, and so the code compiles. It also works since the container isn't empty.
Be aware that there is a difference in this respect between:
auto last = --c.end();
vs.
auto last = c.end();
--last;
The former decrements an rvalue, whereas the latter decrements an lvalue.
You read wrong. The expression --c.end() is never authorized. If the
iterator isn't at least bidirectional, it is, in fact, expressedly
forbidden, and requires a compiler error. If the collection is empty,
it is undefined behavior. And in all other cases, it will work if
it compiles, but there is no guarantee that it will compile. It failed
to compile with many early implementations of std::vector, for
example, where the iterator was just a typedef to a pointer. (In fact,
I think formally that it is undefined behavior in all cases, since
you're violating a constraint on a templated implementation. In
practice, however, you'll get what I just described.)
Arguably, because it isn't guaranteed, a good implementation will cause
it to fail to compile, systematically. For various reasons, most don't.
Don't ask me why, because it's incredibly simple to get it to fail
systematically: just make the operator-- on the iterator a free
function, rather than a member.
EDIT (additional information):
The fact that it isn't required is probably a large part of the
motivation behind std::next and std::prev in C++11. Of course,
every project I've worked on has had them anyway. The correct way to
write this is:
prev( c.end() );
And of course, the constraints that the iterator be bidirectional or
better, and that the container not be empty, still hold.
Each algorithm will tell you what type of iterator it requires. When a bidirectional iterator is called for, then naturally it will need to support decrementing.
Whether --end is possible depends on whether end == beg.
It's only required for algorithms that require bidirectional iterators.

Why is "!=" used with iterators instead of "<"?

I'm used to writing loops like this:
for (std::size_t index = 0; index < foo.size(); index++)
{
// Do stuff with foo[index].
}
But when I see iterator loops in others' code, they look like this:
for (Foo::Iterator iterator = foo.begin(); iterator != foo.end(); iterator++)
{
// Do stuff with *Iterator.
}
I find the iterator != foo.end() to be offputting. It can also be dangerous if iterator is incremented by more than one.
It seems more "correct" to use iterator < foo.end(), but I never see that in real code. Why not?
All iterators are equality comparable. Only random access iterators are relationally comparable. Input iterators, forward iterators, and bidirectional iterators are not relationally comparable.
Thus, the comparison using != is more generic and flexible than the comparison using <.
There are different categories of iterators because not all ranges of elements have the same access properties. For example,
if you have an iterators into an array (a contiguous sequence of elements), it's trivial to relationally compare them; you just have to compare the indices of the pointed to elements (or the pointers to them, since the iterators likely just contain pointers to the elements);
if you have iterators into a linked list and you want to test whether one iterator is "less than" another iterator, you have to walk the nodes of the linked list from the one iterator until either you reach the other iterator or you reach the end of the list.
The rule is that all operations on an iterator should have constant time complexity (or, at a minimum, sublinear time complexity). You can always perform an equality comparison in constant time since you just have to compare whether the iterators point to the same object. So, all iterators are equality comparable.
Further, you aren't allowed to increment an iterator past the end of the range into which it points. So, if you end up in a scenario where it != foo.end() does not do the same thing as it < foo.end(), you already have undefined behavior because you've iterated past the end of the range.
The same is true for pointers into an array: you aren't allowed to increment a pointer beyond one-past-the-end of the array; a program that does so exhibits undefined behavior. (The same is obviously not true for indices, since indices are just integers.)
Some Standard Library implementations (like the Visual C++ Standard Library implementation) have helpful debug code that will raise an assertion when you do something illegal with an iterator like this.
Short answer: Because Iterator is not a number, it's an object.
Longer answer: There are more collections than linear arrays. Trees and hashes, for example, don't really lend themselves to "this index is before this other index". For a tree, two indices that live on separate branches, for example. Or, any two indices in a hash -- they have no order at all, so any order you impose on them is arbitrary.
You don't have to worry about "missing" End(). It is also not a number, it is an object that represents the end of the collection. It doesn't make sense to have an iterator that goes past it, and indeed it cannot.

Why is comparing against "end()" iterator legal?

According to C++ standard (3.7.3.2/4) using (not only dereferencing, but also copying, casting, whatever else) an invalid pointer is undefined behavior (in case of doubt also see this question). Now the typical code to traverse an STL containter looks like this:
std::vector<int> toTraverse;
//populate the vector
for( std::vector<int>::iterator it = toTraverse.begin(); it != toTraverse.end(); ++it ) {
//process( *it );
}
std::vector::end() is an iterator onto the hypothetic element beyond the last element of the containter. There's no element there, therefore using a pointer through that iterator is undefined behavior.
Now how does the != end() work then? I mean in order to do the comparison an iterator needs to be constructed wrapping an invalid address and then that invalid address will have to be used in a comparison which again is undefined behavior. Is such comparison legal and why?
The only requirement for end() is that ++(--end()) == end(). The end() could simply be a special state the iterator is in. There is no reason the end() iterator has to correspond to a pointer of any kind.
Besides, even if it were a pointer, comparing two pointers doesn't require any sort of dereference anyway. Consider the following:
char[5] a = {'a', 'b', 'c', 'd', 'e'};
char* end = a+5;
for (char* it = a; it != a+5; ++it);
That code will work just fine, and it mirrors your vector code.
You're right that an invalid pointer can't be used, but you're wrong that a pointer to an element one past the last element in an array is an invalid pointer - it's valid.
The C standard, section 6.5.6.8 says that it's well defined and valid:
...if the expression P points to the
last element of an array object, the
expression (P)+1 points one past the
last element of the array object...
but cannot be dereferenced:
...if the result points one past the
last element of the array object, it
shall not be used as the operand of a
unary * operator that is evaluated...
One past the end is not an invalid value (neither with regular arrays or iterators). You can't dereference it but it can be used for comparisons.
std::vector<X>::iterator it;
This is a singular iterator. You can only assign a valid iterator to it.
std::vector<X>::iterator it = vec.end();
This is a perfectly valid iterator. You can't dereference it but you can use it for comparisons and decrement it (assuming the container has a sufficient size).
Huh? There's no rule that says that iterators need to be implemented using nothing but a pointer.
It could have a boolean flag in there, which gets set when the increment operation sees that it passes the end of the valid data, for instance.
The implementation of a standard library's container's end() iterator is, well, implementation-defined, so the implementation can play tricks it knows the platform to support.
If you implemented your own iterators, you can do whatever you want - so long as it is standard-conform. For example, your iterator, if storing a pointer, could store a NULL pointer to indicate an end iterator. Or it could contain a boolean flag or whatnot.
I answer here since other answers are now out-of-date; nevertheless, they were not quite right to the question.
First, C++14 has changed the rules mentioned in the question. Indirection through an invalid pointer value or passing an invalid pointer value to a deallocation function are still undefined, but other operations are now implemenatation-defined, see Documentation of "invalid pointer value" conversion in C++ implementations.
Second, words matter. You can't bypass the definitions while applying the rules. The key point here is the definition of "invalid". For iterators, this is defined in [iterator.requirements]. Though pointers are iterators, meanings of "invalid" to them are subtly different. Rules for pointers render "invalid" as "don't indirect through invalid value", which is a special case of "not dereferenceable" to iterators; however, "not deferenceable" is not implying "invalid" for iterators. "Invalid" is explicitly defined as "may be singular", while "singular" value is defined as "not associated with any sequence" (in the same paragraph of definition of "dereferenceable"). That paragraph even explicitly defined "past-the-end values".
From the text of the standard in [iterator.requirements], it is clear that:
Past-the-end values are not assumed to be dereferenceable (at least by the standard library), as the standard states.
Dereferenceable values are not singular, since they are associated with sequence.
Past-the-end values are not singular, since they are associated with sequence.
An iterator is not invalid if it is definitely not singular (by negation on definition of "invalid iterator"). In other words, if an iterator is associated to a sequence, it is not invalid.
Value of end() is a past-the-end value, which is associated with a sequence before it is invalidated. So it is actually valid by definition. Even with misconception on "invalid" literally, the rules of pointers are not applicable here.
The rules allowing == comparison on such values are in input iterator requirements, which is inherited by some other category of iterators (forward, bidirectional, etc). More specifically, valid iterators are required to be comparable in the domain of the iterator in such way (==). Further, forward iterator requirements specifies the domain is over the underlying sequence. And container requirements specifies the iterator and const_iterator member types in any iterator category meets forward iterator requirements. Thus, == on end() and iterator over same container is required to be well-defined. As a standard container, vector<int> also obey the requirements. That's the whole story.
Third, even when end() is a pointer value (this is likely to happen with optimized implementation of iterator of vector instance), the rules in the question are still not applicable. The reason is mentioned above (and in some other answers): "invalid" is concerned with *(indirect through), not comparison. One-past-end value is explicitly allowed to be compared in specified ways by the standard. Also note ISO C++ is not ISO C, they also subtly mismatches (e.g. for < on pointer values not in the same array, unspecified vs. undefined), though they have similar rules here.
Simple. Iterators aren't (necessarily) pointers.
They have some similarities (i.e. you can dereference them), but that's about it.
Besides what was already said (iterators need not be pointers), I'd like to point out the rule you cite
According to C++ standard (3.7.3.2/4)
using (not only dereferencing, but
also copying, casting, whatever else)
an invalid pointer is undefined
behavior
wouldn't apply to end() iterator anyway. Basically, when you have an array, all the pointers to its elements, plus one pointer past-the-end, plus one pointer before the start of the array, are valid. That means:
int arr[5];
int *p=0;
p==arr+4; // OK
p==arr+5; // past-the-end, but OK
p==arr-1; // also OK
p==arr+123456; // not OK, according to your rule