i want to remove elements within a container(for now it is unordered_set) by certain condition
for (auto it = windows.begin(); it != windows.end(); ) {
if ((*it)->closed() == 0)
it = numbers.erase(it);
else
++it;
}
i know the erase(it) will return the position immediately following the last of the elements erased. but
Is it mandatory by the standard there won't cause the rearrangement for the iteation when invoking erase? Is it always safe for all containers and all platforms? Say there may be some magic implementation for certain type of container within certain platform.
The C++ standard requires that unordered_set::erase preserve the order of remaining elements, and return an iterator immediately following those being erased. Therefore, the loop you show is well-defined.
[unord.req]/14 ... The erase members shall invalidate only iterators and references to the erased elements, and preserve the relative order of the elements that are not erased.
[unord.req]/11, Table 91 a.erase(q) Erases the element pointed to by q. Returns the iterator immediately following q prior to the erasure.
For example:
#include <unordered_map>
class A{};
std::unordered_map<unsigned int, A> map {{0,{}},{1, {}},{2, {}}};
int main() {
A& a1 = map[1];
// some insert and remove operations ( key 1 never removed)
// ....
}
Is it safe to still use a1 to reference the value which key is "1", after a lot of insert operations?
In other word:
since std::vector will move elements if the capacity changed, a reference of it's element is not guarantee to be valid. Is this fact also fits for unordered_map?
Yes, it is safe. From the standard:
22.2.7 Unordered associative containers [unord.req]
The insert and emplace members shall not affect the validity of
references to container elements, but may invalidate all iterators to
the container. The erase members shall invalidate only iterators and
references to the erased elements, and preserve the relative order of
the elements that are not erased.
References are safe, but iterators are not!
Perhaps a "less authoritative" source, but easier to read:
References to elements in the unordered_map container remain valid in
all cases, even after a rehash
https://www.cplusplus.com/reference/unordered_map/unordered_map/operator[]/
after a lot of insert operations?
If you meant insert or emplace, then it's fine.
(emphasis mine)
If rehashing occurs due to the insertion, all iterators are invalidated. Otherwise iterators are not affected. References are not invalidated. Rehashing occurs only if the new number of elements is greater than max_load_factor()*bucket_count(). If the insertion is successful, pointers and references to the element obtained while it is held in the node handle are invalidated, and pointers and references obtained to that element before it was extracted become valid. (since C++17)
A std::maps iterators stay valid when inserting elements, eg:
std::map<std::string,int> my_map;
my_map["foo"] = 1;
my_map["bar"] = 2;
auto it_foo = my_map.find("foo");
auto it_bar = my_map.find("bar");
my_map["foobar"] = 3;
after inserting another element (in the last line) the two iterators are still valid. How about the end ? For example:
auto it_end = my_map.find("something that isnt in the map");
my_map["barfoo"] = 4; // does not invalidate iterators
assert(it_end == my_map.end()); // ??
In other words: If a method does not invalidate iterators (other than those explicitly mentioned, as for example in case of map::erase) does this mean that also the end is guaranteed to be the same before as after calling the method?
PS: I am aware that I could just try and see, but this wont tell me whether I can rely on this behaviour.
PPS: For example pushing into a std::vector invalidates all iterators, or only the end (when no reallocation took place), but in this case the docs explicitly mention the end. Following this reasoning, "no iterators are invalidated" should include end, but I am not 100% convinced ;)
N4140 23.2.4 Associative containers [associative.reqmts][1]
9 The insert and emplace 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.
Definitely the term iterators refers to all iterators including end.
I'm using a deque in one of my C++ programs, and was reading the documentation for insert on cppreference.com. Most of it made sense except this bit:
All iterators, including the past-the-end iterator, are invalidated.
References are invalidated too, unless pos == begin() or pos == end(),
in which case they are not invalidated.
What does this mean? Is this saying that references to the deque itself are invalidated, or references to its elements, or references to the iterators? Or something else entirely?
Here's a link the doc in question: http://en.cppreference.com/w/cpp/container/deque/insert
A deque is an object. It is a container, therefore it contains other objects within it's internal storage. Those are elements stored in a deque.
You can access those elements. Accessing an element is basically getting a reference back from the container. If you check it, all of the methods under element access section return a reference type.
You can make a copy of accessed element, but you can store the reference itself. T foo = d.front(); vs T& bar = d.front();. (Let d be some std::deque<T>)
A reference to a deque would be auto& ref_d = &d;. This is something else.
So:
1. What effect does “insert” have on references to deques?
None. references to d are fine.
2. What does this mean?
A deque is designed in such a way that inserting at the beginning or at the end of it does not invalidate the references to the elements which you might have already stored. Though if you insert in the middle, the elements might move in memory. Note that the bar is not touched. Precisely because it cannot be, it gets invalidated. Previously obtained reference (or iterator) doesn't point to anything meaningful anymore, thus dereferencing it is illegal.
3. Is this saying that references to the deque itself are invalidated?
Nope, as in 1.
4. or references to its elements [are invalidated]?
Yes, as in 2.
5. or references to the iterators [are invalidated]?
You again seem to confuse what is what. A reference to an iterator would be std::deque<T>::iterator& iter_ref;, if you obtain an iterator from a deque. E.g. auto iter = d.begin(); and make a reference to it iter_ref = &iter;, an insert doesn't make *iter_ref illegal, it invalidates the iterator, so *iter is illegal (or **ref_iter).
Note: I am not saying that something like std::deque<T>& ref_d or std::deque<T>::iterator& iter_ref make sense, but this is semantical meaning of "reference to a deque" and "reference to an interator".
std::vector<int> ints;
// ... fill ints with random values
for(std::vector<int>::iterator it = ints.begin(); it != ints.end(); )
{
if(*it < 10)
{
*it = ints.back();
ints.pop_back();
continue;
}
it++;
}
This code is not working because when pop_back() is called, it is invalidated. But I don't find any doc talking about invalidation of iterators in std::vector::pop_back().
Do you have some links about that?
The call to pop_back() removes the last element in the vector and so the iterator to that element is invalidated. The pop_back() call does not invalidate iterators to items before the last element, only reallocation will do that. From Josuttis' "C++ Standard Library Reference":
Inserting or removing elements
invalidates references, pointers, and
iterators that refer to the following
element. If an insertion causes
reallocation, it invalidates all
references, iterators, and pointers.
Here is your answer, directly from The Holy Standard:
23.2.4.2 A vector satisfies all of the requirements of a container and of a reversible container (given in two tables in 23.1) and of a sequence, including most of the optional sequence requirements (23.1.1).
23.1.1.12 Table 68
expressiona.pop_back()
return typevoid
operational semanticsa.erase(--a.end())
containervector, list, deque
Notice that a.pop_back is equivalent to a.erase(--a.end()). Looking at vector's specifics on erase:
23.2.4.3.3 - iterator erase(iterator position) - effects - Invalidates all the iterators and references after the point of the erase
Therefore, once you call pop_back, any iterators to the previously final element (which now no longer exists) are invalidated.
Looking at your code, the problem is that when you remove the final element and the list becomes empty, you still increment it and walk off the end of the list.
(I use the numbering scheme as used in the C++0x working draft, obtainable here
Table 94 at page 732 says that pop_back (if it exists in a sequence container) has the following effect:
{ iterator tmp = a.end();
--tmp;
a.erase(tmp); }
23.1.1, point 12 states that:
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.
Both accessing end() as applying prefix-- have no such effect, erase() however:
23.2.6.4 (concerning vector.erase() point 4):
Effects: Invalidates iterators and references at or after the point of the erase.
So in conclusion: pop_back() will only invalidate an iterator to the last element, per the standard.
Here is a quote from SGI's STL documentation (http://www.sgi.com/tech/stl/Vector.html):
[5] A vector's iterators are invalidated when its memory is reallocated. Additionally, inserting or deleting an element in the middle of a vector invalidates all iterators that point to elements following the insertion or deletion point. It follows that you can prevent a vector's iterators from being invalidated if you use reserve() to preallocate as much memory as the vector will ever use, and if all insertions and deletions are at the vector's end.
I think it follows that pop_back only invalidates the iterator pointing at the last element and the end() iterator. We really need to see the data for which the code fails, as well as the manner in which it fails to decide what's going on. As far as I can tell, the code should work - the usual problem in such code is that removal of element and ++ on iterator happen in the same iteration, the way #mikhaild points out. However, in this code it's not the case: it++ does not happen when pop_back is called.
Something bad may still happen when it is pointing to the last element, and the last element is less than 10. We're now comparing an invalidated it and end(). It may still work, but no guarantees can be made.
Iterators are only invalidated on reallocation of storage. Google is your friend: see footnote 5.
Your code is not working for other reasons.
pop_back() invalidates only iterators that point to the last element. From C++ Standard Library Reference:
Inserting or removing elements
invalidates references, pointers, and
iterators that refer to the following
element. If an insertion causes
reallocation, it invalidates all
references, iterators, and pointers.
So to answer your question, no it does not invalidate all iterators.
However, in your code example, it can invalidate it when it is pointing to the last element and the value is below 10. In which case Visual Studio debug STL will mark iterator as invalidated, and further check for it not being equal to end() will show an assert.
If iterators are implemented as pure pointers (as they would in probably all non-debug STL vector cases), your code should just work. If iterators are more than pointers, then your code does not handle this case of removing the last element correctly.
Error is that when "it" points to the last element of vector and if this element is less than 10, this last element is removed. And now "it" points to ints.end(), next "it++" moves pointer to ints.end()+1, so now "it" running away from ints.end(), and you got infinite loop scanning all your memory :).
The "official specification" is the C++ Standard. If you don't have access to a copy of C++03, you can get the latest draft of C++0x from the Committee's website: http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2008/n2723.pdf
The "Operational Semantics" section of container requirements specifies that pop_back() is equivalent to { iterator i = end(); --i; erase(i); }. the [vector.modifiers] section for erase says "Effects: Invalidates iterators and references at or after the point of the erase."
If you want the intuition argument, pop_back is no-fail (since destruction of value_types in standard containers are not allowed to throw exceptions), so it cannot do any copy or allocation (since they can throw), which means that you can guess that the iterator to the erased element and the end iterator are invalidated, but the remainder are not.
pop_back() will only invalidate it if it was pointing to the last item in the vector. Your code will therefore fail whenever the last int in the vector is less than 10, as follows:
*it = ints.back(); // Set *it to the value it already has
ints.pop_back(); // Invalidate the iterator
continue; // Loop round and access the invalid iterator
You might want to consider using the return value of erase instead of swapping the back element to the deleted position an popping back. For sequences erase returns an iterator pointing the the element one beyond the element being deleted. Note that this method may cause more copying than your original algorithm.
for(std::vector<int>::iterator it = ints.begin(); it != ints.end(); )
{
if(*it < 10)
it = ints.erase( it );
else
++it;
}
std::remove_if could also be an alternative solution.
struct LessThanTen { bool operator()( int n ) { return n < 10; } };
ints.erase( std::remove_if( ints.begin(), ints.end(), LessThanTen() ), ints.end() );
std::remove_if is (like my first algorithm) stable, so it may not be the most efficient way of doing this, but it is succinct.
Check out the information here (cplusplus.com):
Delete last element
Removes the last element in the vector, effectively reducing the vector size by one and invalidating all iterators and references to it.