Say I have something like this:
void myFunk(std::vector<T>& v, std::vector<T>::iterator first, std::vector<T>::iterator last) {
while (first != last) {
if ((*first) > (*last)) {
T someT;
v.push_back(someT);
}
first++;
}
}
int main(){
std::vector<T> foo = {some, T, values};
myFunky(foo, foo.begin(), foo.end())
return 0;
}
Would this lead to an infinite loop, or would it end after foo.size() iterations? In other words, would the last iterator be updated as foo grew, or would it retain the value given in the function call?
I'm assuming last would change, since it's a pointer to a position, but would like some confirmation.
Would this lead to an infinite loop, or would it end after foo.size() iterations?
Neither. What you are doing is undefined behavior, for a couple of reasons:
You are modifying the vector while iterating through it.
If the vector reallocates its internal storage when pushing a new item, all existing iterators into the vector are invalidated, including both iterators you are using to loop with. But even just pushing a new item always invalidates the end() iterator, at least.
See Iterator invalidation rules for C++ containers
You are dereferencing the end() iterator, which never refers to a valid element.
I'm assuming last would change, since it's a pointer to a position
It can't change, since you passed it into the myFunc function by value, so it is a copy of the original end() iterator. If end() changes value, last will not change value, since it is a copy.
In any case, iterators are not necessarily implemented as pointers, but pointers are valid iterators. But it doesn't matter in this case. Even if vector::iterator were just a simple pointer, last would still get invalidated upon every push/reallocation.
Related
Following lines of C++ code gives runtime error but if erase operation mymap.erase(v) is removed it works:
map<int,int> mymap = {{1,0},{2,1},{9,2},{10,3},{11,4}};
for(auto it=mymap.rbegin();it!=mymap.rend();){
int v=it->first;
++it;
mymap.erase(v);
}
demo
Here iterator it is changed before deleting its value v, so iterator it should remain unaffected I believe.
When you are calling erase(v), you are invalidating the base iterator that the next reverse_iterator (from ++it) is using. So you need to create a new reverse_iterator from the base iterator that precedes the erased value.
Also, rather than erasing the value that the reverse_iterator is referring to, you should erase the base iterator instead, since you already know which element you want to erase. There is no need to make the map go hunting for the value again.
This works for me:
map<int,int> mymap = {{1,0},{2,1},{9,2},{10,3},{11,4}};
for(auto it = mymap.rbegin(); it != mymap.rend(); ){
auto v = --(it.base());
v = mymap.erase(v);
it = map<int,int>::reverse_iterator(v);
}
Demo
On the other hand, this loop is essentially just erase()'ing all elements from mymap, so a better option is to use mymap.clear() instead.
Indeed, std::map::erase:
References and iterators to the erased elements are invalidated. Other references and iterators are not affected.
But std::reverse_iterator
For a reverse iterator r constructed from an iterator i, the relationship &*r == &*(i-1) is always true (as long as r is dereferenceable); thus a reverse iterator constructed from a one-past-the-end iterator dereferences to the last element in a sequence.
Perhaps it gets more clear when you look at the image on the cppreference page.
The crucial part is "Reverse iterator stores an iterator to the next element than the one it actually refers to".
As a consequence (small modification to your code)
auto& element = *it; // fails in the next iteration, because...
int v = element.first;
++it; // now it stores an iterator to element
mymap.erase(v); // iterators to element are invalidated
You are erasing the element that is used by it in the next iteration.
Because vectors use an array as their underlying storage, inserting
elements in positions other than the vector end causes the container
to relocate all the elements that were after position to their new
positions.
< http://www.cplusplus.com/reference/vector/vector/insert/ >
I thought that this is the reason that iterator it becomes no longer valid after the last line in code below:
std::vector<int> myvector (3,100);
std::vector<int>::iterator it;
it = myvector.begin();
it = myvector.insert ( it , 200 );
myvector.insert (it,2,300);
But if I change the it's definition into myvector.end();, it's still the same. What is the reason behind this? How exactly does it work and are there situations where iterator insert can be still valid after filling part of vector with some elements? (or single one)
Well yes, that is a reason for iterators to elements after (and at, because insertion is done before the given element) the insertion point are invalidated. If you insert to the end, then there are no elements whose iterators could be invalidated. The end iterator is always invalidated, no matter where you insert. The more relevant description on that page:
Iterator validity
If a reallocation happens, all iterators, pointers and references related to the container are invalidated.
Otherwise, only those pointing to position and beyond are invalidated, with all iterators, pointers and references to elements before position guaranteed to keep referring to the same elements they were referring to before the call.
Here's what it points to if you change the first assignment to end.
it = myvector.end();
it points to end, good.
it = myvector.insert ( it , 200 );
Inserting to end does not invalidate any pointers to elements, but it does invalidate the end iterator which is the old value for it. Luckily, you now assign to the iterator returned by insert. That iterator does not point to the end of the vector but to the newly inserted element.
myvector.insert (it,2,300);
Now it is invalidated again, but you don't reassign it, so it remains so.
Of course, then there is the possibility, after each insert, that the vector was reallocated in which case all previous iterators to any part of the vector would be invalidated. That can be avoided by guaranteeing sufficient space with vector::reserve before initializing the iterators. The new iterator returned by insert will always be valid, even if the vector was reallocated.
Here is a better reference and explanation.
Causes reallocation if the new size() is greater than the old capacity(). If the new size() is greater than capacity(), all iterators and references are invalidated. Otherwise, only the iterators and references before the insertion point remain valid. The past-the-end iterator is also invalidated.
— http://en.cppreference.com/w/cpp/container/vector/insert
In the case where you use end(), size() still goes above capacity(). Try setting the capacity to something larger before insert().
STL iterators are used with container classes and are conceptually similar to pointers to specific elements stored in the container.
One of the statements below is true. Which one?
An iterator typically holds an address (pointer), and operator++ applied to the iterator always increases that address.
When iterator it goes out of scope in a program, it gets destructed, which automatically invokes delete it;.
For a valid STL container myC, when the expression myC.end()-myC.begin() is well-defined, it returns the same value as myC.size().
When a container goes out of scope, all iterators that point to it are automatically modified.
For a valid STL container myC, the iterator returned by myC.end() refers to the last valid element in myC.
Apparently the solution is 3. but I don't understand why. Can someone elaborate on why this is the case, and possibly show why the others are false as well?
Think of the requirements of the addresses of items in a linked-list (list). They don't need to be sequential in memory.
delete is something that's manually done on pointers, it wouldn't happen automatically (even if the pointer goes out of scope) (unless done in some API). Iterators are (generally) classes, so delete would not even apply. The iterator would get destructed though.
You can also probably classify a pointer as an iterator. But delete will still not be called automatically.
Note that this only applies to random access iterators. You can calculate the number of items in a container as follows:
int count = 0;
for (iterator it = begin(); it != end(); ++it, ++count) { }
so you increment begin() count times to get to end(),
so begin() + count = end(),
so end() - begin() = count, and count = size(),
so end() - begin() = size()
This is not the way C++ works. Although there are design patterns to achieve this behaviour, usually when modifying a class, it's your responsibility to ensure any dependent classes are updated if invalidated. When you'd try to use an iterator of a container that went out of scope, this would result in undefined behaviour.
end() is past the last element, probably with something like this in mind: (I'm sure among other reasons)
for (iterator it = begin(); it != end(); ++it)
I'm curious about the rationale behind the following code. For a given map, I can delete a range up to, but not including, end() (obviously,) using the following code:
map<string, int> myMap;
myMap["one"] = 1;
myMap["two"] = 2;
myMap["three"] = 3;
map<string, int>::iterator it = myMap.find("two");
myMap.erase( it, myMap.end() );
This erases the last two items using the range. However, if I used the single iterator version of erase, I half expected passing myMap.end() to result in no action as the iterator was clearly at the end of the collection. This is as distinct from a corrupt or invalid iterator which would clearly lead to undefined behaviour.
However, when I do this:
myMap.erase( myMap.end() );
I simply get a segmentation fault. I wouldn't have thought it difficult for map to check whether the iterator equalled end() and not take action in that case. Is there some subtle reason for this that I'm missing? I noticed that even this works:
myMap.erase( myMap.end(), myMap.end() );
(i.e. does nothing)
The reason I ask is that I have some code which receives a valid iterator to the collection (but which could be end()) and I wanted to simply pass this into erase rather than having to check first like this:
if ( it != myMap.end() )
myMap.erase( it );
which seems a bit clunky to me. The alternative is to re code so I can use the by-key-type erase overload but I'd rather not re-write too much if I can help it.
The key is that in the standard library ranges determined by two iterators are half-opened ranges. In math notation [a,b) They include the first but not the last iterator (if both are the same, the range is empty). At the same time, end() returns an iterator that is one beyond the last element, which perfectly matches the half-open range notation.
When you use the range version of erase it will never try to delete the element referenced by the last iterator. Consider a modified example:
map<int,int> m;
for (int i = 0; i < 5; ++i)
m[i] = i;
m.erase( m.find(1), m.find(4) );
At the end of the execution the map will hold two keys 0 and 4. Note that the element referred by the second iterator was not erased from the container.
On the other hand, the single iterator operation will erase the element referenced by the iterator. If the code above was changed to:
for (int i = 1; i <= 4; ++i )
m.erase( m.find(i) );
The element with key 4 will be deleted. In your case you will attempt to delete the end iterator that does not refer to a valid object.
I wouldn't have thought it difficult for map to check whether the iterator equalled end() and not take action in that case.
No, it is not hard to do, but the function was designed with a different contract in mind: the caller must pass in an iterator into an element in the container. Part of the reason for this is that in C++ most of the features are designed so that the incur the minimum cost possible, allowing the user to balance the safety/performance on their side. The user can test the iterator before calling erase, but if that test was inside the library then the user would not be able to opt out of testing when she knows that the iterator is valid.
n3337 23.2.4 Table 102
a.erase( q1, q2)
erases all the elements in the range [q1,q2). Returns q2.
So, iterator returning from map::end() is not in range in case of myMap.erase(myMap.end(), myMap.end());
a.erase(q)
erases the element pointed to by q. Returns an iterator pointing to the element immediately following q prior to the element being erased. If no such element exists, returns a.end().
I wouldn't have thought it difficult for map to check whether the
iterator equalled end() and not take action in that case. Is there
some subtle reason for this that I'm missing?
Reason is same, that std::vector::operator[] can don't check, that index is in range, of course.
When you use two iterators to specify a range, the range consists of the elements from the element that the first iterator points to up to but not including the element that the second iterator points to. So erase(it, myMap.end()) says to erase everything from it up to but not including end(). You could equally well pass an iterator that points to a "real" element as the second one, and the element that that iterator points to would not be erased.
When you use erase(it) it says to erase the element that it points to. The end() iterator does not point to a valid element, so erase(end()) doesn't do anything sensible. It would be possible for the library to diagnose this situation, and a debugging library will do that, but it imposes a cost on every call to erase to check what the iterator points to. The standard library doesn't impose that cost on users. You're on your own. <g>
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.