Is it possible to peek next element in a container which the iterator currently points to without changing the iterator?
For example in std::set,
int myArray[]= {1,2,3,4};
set <int> mySet(myArray, myArray+4);
set <int>::iterator iter = mySet.begin();
//peek the next element in set without changing iterator.
mySet.erase(iter); //erase the element if next element is n+1
C++0x adds a handy utility function, std::next, that copies an iterator, advances it, and returns the advanced iterator. You can easily write your own std::next implementation:
#include <iterator>
template <typename ForwardIt>
ForwardIt next(ForwardIt it,
typename std::iterator_traits<ForwardIt>::difference_type n = 1)
{
std::advance(it, n);
return it;
}
You can use this in your example like so:
if (iter != mySet.end() && next(iter) != mySet.end() && *next(iter) == *iter + 1)
mySet.erase(iter);
Not with iterators in general. An iterator isn't guaranteed to be able to operate non-destructively. The classic example is an Input Iterator that actually represents an underlying input stream.
There's something that works for this kind of iterator, though. A Forward Iterator doesn't invalidate previous copies of itself by the act of moving forward through the collection. Most iterators (including those for STL collections) are at least Forward Iterators, if not a more functional version- only Input Iterators or Output Iterators are more restricted. So you can simply make a copy of your iterator, increment the copy and check that, then go back to your original iterator.
So your peek code:
set <int>::iterator dupe = iter;
++dupe;
// (do stuff with dupe)
set <int>::iterator iter2 = iter;
++iter2;
int peekedValue = *iter2;
You can always make a copy of the iterator and advance the copy:
set <int>::iterator iter = mySet.begin();
set <int>::iterator iterCopy = iter;
iterCopy++;
if (*iterCopy == something)
mySet.erase(iter);
But beware that iterCopy may no longer be valid once you erase iter.
for sequence containers (vector, deque, and list) you can call front which will give you a peek (more info on the lower part of this link).
This will not work for std::set as its nature does not allow for the [] operator, but for containers that do, you can do:
std::vector<int> v;
v.push_back(3);
v.push_back(4);
std::vector<int>::iterator it = v.begin();
std::cout << v[it - v.begin() + 1];
But this could be dangerous if it points to the last element in the container; but the same applies to the solution above. E.g. you'll have to make checks in both cases.
Related
Since we now have advance() and the prev() to move iterator to go front or go back, and we already have begin() and end().
I wonder is there any situation we better/have to move reverse iterator back and front?
Algorithms often take two iterators that specify a range of elements. For example std::for_each:
std::vector<int> x;
std::for_each(x.begin(),x.end(),foo);
If you want to make for_each iterate in reverse order (note: for_each does iterate in order) then neither advance nor prev are of any help, but you can use reverse iterators:
std::for_each(x.rbegin(),x.rend(),foo);
Because using begin() and end() to iterate in reverse looks horrible:
std::vector<int> v {1, 2, 3};
if(!v.empty()) { //need to make sure of that before we decrement
for(auto it = std::prev(v.end()); ; --it) {
//do something with it
if(it == v.begin()) {
break;
}
}
}
Compare it with reverse iterator version:
std::vector<int> v {1, 2, 3};
for(auto it = v.rbegin(); it != v.rend(); it++) {
//do something with it
}
When you have a function template that takes iterators, and want it to operate on the data in reverse.
E.g.
std::string s = "Hello";
std::string r(s.rbegin(), s.rend());
std::cout << r;
When you use algorithms like std::for_each(), std::accumulate(), std::find_if()... they systematically progress with ++.
If you want this progression to physically occur backwards, then the reverse
iterators are useful.
I guess it is good practise because it seems odd if you start from end and finish in begin. You can easily say last but one by using rbegin.
vector::reverse_iterator itr1;
for (itr1 = vec.rbegin(); itr1 < vec.rend(); itr1++) {
if (*itr1 == num) {
vec.erase((itr1 + 1).base());
}
}
You can use as a function which deletes that Which num want to erase in vector
The need for rbegin()/rend() is because begin() is not the same as rend(), and end() is not rbegin(), see this image from cppreference
This way, you can use any algorithm going forward from beginning to end or backwards from the last to the first element.
There are examples with for each. However, more general, it allows you to reuse any algorithm or operators that works with iterators with advancing, to do the same thing but in a reverse order.
I was about to write code like this:
std::list<whatevertype> mylist;
// ...
std::list<whatevertype>::iterator it;
for(it = mylist.begin(); it != mylist.end(); ++it) {
// ...
if(some condition)
mylist.erase(it);
}
But I realized, this code is wrong: mylist.erase(x) will invalidate the iterator it, so the ++it is likely to fail.
So I tried changing it to
std::list<whatevertype>::iterator it;
std::list<whatevertype>::iterator nextit;
for(it = mylist.begin(); it != mylist.end(); it = nextit) {
// ...
nextit = it + 1;
if(some condition)
mylist.erase(it);
}
But, to my surprise, this failed: evidently operator+ is not defined for std::list iterators.
I've since found this other question and learned that the standard idiom for deleting "out from under" an iterator is more like
for(it = mylist.begin(); it != mylist.end(); ) {
if(some condition)
it = mylist.erase(it);
else ++it;
}
I believe I could also get away with
for(it = mylist.begin(); it != mylist.end(); ) {
// ...
std::list<whatevertype>::iterator previt = it;
++it;
if(some condition)
mylist.erase(previt);
}
But my question is, is there a reason that operator+ is not defined for these iterators?
One rule they had with the std iterators and collection was to make expensive things verbose.
On a list iterator, it+50 takes O(50) time. On a vector iterator, it+50 takes O(1) time. So they implemented + on vector iterators (and other random access iterators) but not on list iterators (and other weaker iterators).
std::next and std::advance and std::prev can solve your problem easier:
auto previt = std::prev(it);
or
auto nextit = std::next(it);
these also take a count, but because they are an explicit function call it was decided that them being expensive is acceptable.
Among other things, you can search for calls to std::next and std::prev and get iterator manipulation; + is heavily overloaded and finding the expensive calls is hard.
Note that std::basic_string doesn't follow the same conventions as other std containers.
It isn't that + is missing for all iterators. It is missing for std::list iterators.
That's because a list iterator is incredibly inefficient at random access. Therefore, making random access easy is a bad idea.
You can use std::advance. It makes it more evident that you are moving across the list one element at a time.
std::list uses a BidirectionalIterator which only defines increment and decrement. As std::list is a linked list the implementation of the iterator can only move one node at a time.
The interface is designed to make sure you know that moving by more than one element isn't a simple operation like it is with other iterators like a RandomAccessIterator returned from a std::vector.
see http://en.cppreference.com/w/cpp/concept/Iterator for a definition of the different iterator types.
I have a map which elements are vectors.I have to delete from these vectors all elements which are equal to special number num
std::map<size_t,std::vector<size_t> > myMap;
for (std::map<size_t,std::vector<size_t> >::iterator itMap = myMap.begin();itMap != myMap.end();++itMap )
{
for (std::vector<size_t>::iterator itVec = itMap->second.begin();itVec != itMap->second.end();)
{
auto itNextVec = itVec;
++itNextVec;
if (*itVec == num)
{
itMap->second.erase(itVec );
}
itVec = itNextVec;
}
}
The code causes run-time exepssion .In VS - vector iterators incompatible.
Can someone point what is the cause for that?
Thanks
std::vector::erase returns an iterator to the next position of the list, and so when you do an erase you should make your iterator equal to the returned value.
The only thing that you have to consider is that the returned iterator could be the end so you should check for that.
What I personally like to do is is after doing in an erase and I get the next iterator position, I go back to the previous position of the returned iterator and than call a continue on the for loop
Example:
#include <vector>
#include <iostream>
int main()
{
std::vector<int> myInt;
myInt.push_back(1);myInt.push_back(2);myInt.push_back(3);
for(auto iter = myInt.begin();
iter != myInt.end();
++iter)
{
if(*iter == 1)
{
iter = myInt.erase(iter);
if(iter != myInt.begin())
{
iter = std::prev(iter);
continue;
}
}
std::cout << *iter << std::endl;
}
}
But doing an erase inside of a iterator loop is frowned upon because it invalidates the old iterator and that could cause a lot of issues if you didn't plan for them.
erasing will invalidate the iterator
Iterator validity
Iterators, pointers and references pointing to position (or first) and beyond are
invalidated, with all iterators, pointers and references to elements before position (or
first) are guaranteed to keep referring to the same elements they were referring to
before the call.
You can't trivially erase an item from a collection while iterating over it. Think a little about it, your removing what itVec "points" to, after the removal itVec no longer "points" to an element, so it no longer have a "next" pointer.
If you check e.g. this reference, you will see that the erase function returns an iterator to the next element. Continue the loop with this one (without increasing it of course).
Consider either using a different collection class than vector or creating a new vector with the desired items removed rather than removing from existing vector.
For example, it this expression valid in semantic?
container.begin() == container.begin();
Yes, so long as neither iterator has been invalidated.
For example, the following would not be valid:
std::deque<int> d;
std::deque<int> begin1 = d.begin();
d.push_front(42); // invalidates begin1!
std::deque<int> begin2 = d.begin();
assert(begin1 == begin2); // wrong; you can't use begin1 anymore.
Yes, begin() will return the same iterator given a container instance, unless you change the container in some way (end() has this property as well). For example, std::vector::push_back() may cause the array to be reallocated to accommodate new elements.
I was wondering if something like this is safe...
// Iterating through a <list>
while ( iter != seq.end()) {
if ( test ) {
iter = seq.erase( iter );
} else {
++iter;
}
I know that iterating through a vector in this way would invalidate the iterator, but would the same thing occur in a list? I assume not since a list is sequential through pointers rather than being "next" to each other in memory, but any reassurance would be helpful.
This is just fine because the erase method returns a new valid iterator.
Yes -- std::list::erase(): "Invalidates only the iterators and references to the erased elements."
That said, you probably shouldn't do this at all -- you seem to be trying to imitate std::remove_if().
The standard defines erase behaviour for every STL container. For std::list only iterators to the erased elements are invalidated. The return value of erase needn't be a dereferencable one, though (it could be list.end()).
Therefore, to erase all elements in a list the following is absolutely valid:
.. it = l.begin();
while(it != l.end()) {
it = l.erase(it);
}
BUT beware of something like this (dangerous pitfall):
for (.. it = l.begin; it != l.end(); ++it) {
it = l.erase(it);
}
If it is l.end(), it is incremented twice (second time by the loop head). Baamm.
Yes, this is the standard way to do that. See Effective STL, Item 9 (p. 46).
Yes, this is totally safe. The erase() function returns an iterator to the element succeeding the one which was erased. Had you not reassigned the result of erase() to iter, you'd have trouble.
As others have explained, your code does not invalidate the iterator used in the function. However, it does invalidate other iterators if the collection is a vector, but not if the collection is a list.
As others have mentioned, yes, it will work. But I'd recommend using list::remove_if instead, as it's more expressive.