Is it allowed to increment an iterator variable it that already is at end(), i.e. auto it = v.end()?
Is it allowed in general?
If not, is it not allowed for vector?
If yes, is ++it maybe idempotent if it==v.end()?
I ask, because I stumbled upon code like this:
std::vector<int> v{ 1, 2, 3, 4, 5, 6, 7 };
// delete every other element
for(auto it=v.begin(); it<v.end(); ++it) { // it<end ok? ++it ok on end?
it = v.erase(it);
}
It works fine with g++-6, but that is no proof.
For one it<v.end() may only work with vectors, I suppose it should read it!=v.end() in general. But in this example that would not recognize the end of v if ++it is applied when it already is on the end.
No the behaviour is undefined. You are allowed to set an iterator to end(), but you must not increment it or dereference it.
You are allowed to decrement it so long as the backing container is not empty.
Related
Consider the following code
std::set<int> int_set = {1, 2, 3, 4};
for(const auto& key : int_set)
{
if(key == 2)
{
int_set.erase(key);
break;
}
}
The code runs as expected, but is it safe?
It feels wrong to be using a reference to a key to erase itself from a set, as presumably once the erase has happened the reference is no longer valid.
Another code snippet with the same potential problem would be
std::set<int> int_set = {1, 2, 3, 4};
const auto& key = *int_set.find(2);
int_set.erase(k);
This is safe in the code provided since you break out of the loop without attempting to use the reference (key) again (nor implicitly advance the underlying iterator that for-each loops are implemented in terms of, which would happen if you did not break/return/throw/exit()/crash, when you looped back to the top of the loop) after the call to erase. key itself is only needed to find the element to remove; until that element is removed, it's valid, once it's removed, it's not used again by erase (it already found the element to erase, there's no possible use for it after that point).
If you tried to use key after the erase, you'd be using a dangling reference, invoking undefined behavior. Similarly, even allowing the loop to continue (implicitly advancing the underlying iterator) would be illegal; the iterator is invalid, and the implicit advancing of the iterator when you returned to the top of the loop would be equally invalid. The safe way to erase more than one element as you iterate would be to switch from for-each loops (that are convenient, but inflexible) to using iterators directly, so you can update them with the return value of erase, e.g.:
for(auto it = int_set.begin(); it != int_set.end(); /* Don't increment here */)
{
if (predicate(*it)) {
it = int_set.erase(it); // In C++11 and higher, erase returns an iterator to the
// element following the erased element so we can
// seamlessly continue processing
} else {
++it; // Increment if we didn't erase anything
}
}
Of course, as noted in the comments, if you only need to remove one element with a known value, the whole loop is pointless, being a slow (O(n) loop + O(log n) erase) way to spell:
int_set.erase(2); // Returns 1 if 2 was in the set, 0 otherwise
which is a single O(log n) operation.
I am working on an exercise where I have a vector and I am writing my own reverse algorithm by using a reverse and a normal (forward) iterator to reverse the content of the vector. However, I am not able to compare the iterators.
int vals[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 0 };
vector<int> numbers(vals, vals + 10);
vector<int>::iterator start = numbers.begin();
vector<int>::reverse_iterator end = numbers.rend();
I have a previous algorithm for reversing the vector by using two iterators, however in this task I am not able to compare them using the != operator between them. My guess would be to get the underlying pointers or indexes in the vector with each other but how do I get the pointers/index?
Do a comparison using the the iterator returned by base(): it == rit.base() - 1.
You can convert a reverse_iterator to iterator by calling base().
Be careful however, as there are some caveats. #Matthieu M.'s comment is particularly helpful:
Note: base() actually returns an iterator to the element following the
element that the reverse_iterator was pointing to.
Checkout http://en.cppreference.com/w/cpp/iterator/reverse_iterator/base
rit.base()
returns a 'normal' iterator.
You can use (&*start == &*(end - 1)) to directly compare the address that the iterator is pointing to.
The two types cannot be compared (which is a very good idea) and calling .base() is not very elegant (or generic) in my opinion.
You can convert the types and compare the result.
Taking into account the off-by-one rule involving reverse_iterators.
Conversion from iterator to reverse_iterator need to be explicit (fortunately), however, conversion from reverse_iterator to iterator is not possible (unfortunately).
So there is only one way to do conversion and then make the comparison.
std::vector<double> vv = {1.,2.,3.};
auto it = vv.begin();
auto rit = vv.rend();
// assert( it == rit ); // error: does not compile
assert(std::vector<double>::reverse_iterator{it} == rit);
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 want to convert a vector into a sub-range of that vector, for example by removing the first and last values. Is the use of the assign member function valid in this context?
std::vector<int> data = {1, 2, 3, 4};
data.assign(data.begin() + 1, data.end() - 1);
// data is hopefully {2, 3}
Cppreference states that
All iterators, pointers and references to the elements of the container are invalidated. The past-the-end iterator is also invalidated.
This invalidation, however, doesn't appear to happen until the end of assign.
To be safe, I could just go with the following but it seems more verbose:
std::vector<int> data = {1, 2, 3, 4};
data = std::vector<int>{data.begin() + 1, data.end() - 1};
// data is now {2, 3}
The __invalidate_all_iterators function that your link refers to is merely a debugging tool. It doesn't "cause" the iterators to be invalidated; It effectively reports that the iterators have been invalidated by the previous actions. It may be that this debugging tool might not catch a bug caused by this assignment.
It is a precondition of assign that the iterators are not to the same container. A precondition violation results in undefined behaviour.
Standard quote (latest draft):
[sequence.reqmts] a.assign(i,j) Expects: T is Cpp17EmplaceConstructible into X from *i and assignable from *i.
For vector, if the iterator does not meet the forward iterator requirements ([forward.iterators]), T is also Cpp17MoveInsertable into X.
Neither i nor j are iterators into a.
Your safe alternative is correct.
If you want to avoid reallocation (keeping in mind that there will be unused space left), and if you want to avoid copying (which is important for complex types, doesn't matter for int), then following should be efficient:
int begin_offset = 1;
int end_offset = 1;
if (begin_offset)
std::move(data.begin() + begin_offset, data.end() - end_offset, data.begin());
data.erase(data.end() - end_offset - begin_offset, data.end());
I am bit confused about the difference between the usage of std::remove algorithm. Specifically I am not able to understand what is being removed when I use this algorithm. I wrote a small test code like this:
std::vector<int> a;
a.push_back(1);
a.push_back(2);
std::remove(a.begin(), a.end(), 1);
int s = a.size();
std::vector<int>::iterator iter = a.begin();
std::vector<int>::iterator endIter = a.end();
std::cout<<"Using iter...\n";
for(; iter != endIter; ++iter)
{
std::cout<<*iter<<"\n";
}
std::cout<<"Using size...\n";
for(int i = 0; i < a.size(); ++i)
{
std::cout<<a[i]<<"\n";
}
The output was 2,2 in both the cases.
However, if I use erase with the remove something like this:
a.erase(std::remove(a.begin(), a.end(), 1), a.end());
I get the output as 2.
So my questions are:
(1). Is there any use of std::remove other than using it with erase function.
(2). Even after doing std::remove, why a.size() returns 2 and not 1?
I read the item in Scott Meyer's Effective STL book about the erase-remove idiom. But am still having this confusion.
remove() doesn't actually delete elements from the container -- it only shunts non-deleted elements forwards on top of deleted elements. The key is to realise that remove() is designed to work on not just a container but on any arbitrary forward iterator pair: that means it can't actually delete the elements, because an arbitrary iterator pair doesn't necessarily have the ability to delete elements.
For example, pointers to the beginning and end of a regular C array are forward iterators and as such can be used with remove():
int foo[100];
...
remove(foo, foo + 100, 42); // Remove all elements equal to 42
Here it's obvious that remove() cannot resize the array!
What does std::remove do?
Here's pseudo code of std::remove. Take few seconds to see what it's doing and then read the explanation.
Iter remove(Iter start, Iter end, T val) {
Iter destination = start;
//loop through entire list
while(start != end) {
//skip element(s) to be removed
if (*start == val) {
start++;
}
else //retain rest of the elements
*destination++ = *start++;
}
//return the new end of the list
return destination;
}
Notice that remove simply moved up the elements in the sequence, overwriting the values that you wanted to remove. So the values you wanted to remove are indeed gone, but then what's the problem? Let say you had vector with values {1, 2, 3, 4, 5}. After you call remove for val = 3, the vector now has {1, 2, 4, 5, 5}. That is, 4 and 5 got moved up so that 3 is gone from the vector but the size of vector hasn't changed. Also, the end of the vector now contains additional left over copy of 5.
What does vector::erase do?
std::erase takes start and end of the range you want to get rid off. It does not take the value you want to remove, only start and end of the range. Here's pseudo code for how it works:
erase(Iter first, Iter last)
{
//copy remaining elements from last
while (last != end())
*first++ = *last++;
//truncate vector
resize(first - begin());
}
So the erase operation actually changes the size of container and frees up the memory.
The remove-erase idiom
The combination of std::remove and std::erase allows you to remove matching elements from the container so that container would actually get truncated if elements were removed. Here's how to do it:
//first do the remove
auto removed = std::remove(vec.begin(), vec.end(), val);
//now truncate the vector
vec.erase(removed, vec.end());
This is known as the remove-erase idiom. Why is it designed like this? The insight is that the operation of finding elements is more generic and independent of underlying container (only dependent on iterators). However operation of erase depends on how container is storing memory (for example, you might have linked list instead of dynamic array). So STL expects containers to do its own erase while providing generic "remove" operation so all containers don't have to implement that code. In my view, the name is very misleading and std::remove should have been called std::find_move.
Note: Above code is strictly pseudocode. The actual STL implementation is more smarter, for example, using std::move instead of copy.
std::remove does not remove the actual objects, rather, pushes them to the end of the container. Actual deletion and deallocation of memory is done via erase. So:
(1). Is there any use of std::remove other than using it with erase function.
Yes, it helps to get a pair of iterators to a new sequence without having worry about proper de-allocation etc.
(2). Even after doing std::remove, why a.size() returns 2 and not 1?
The container still holds to those objects, you only have a new set of iterators to work with. Hence the size is still what it used to be.
i faced the same issue, trying to understand the difference.
the explanations that have been give so far are right on the money, but i only understood them after seeing an example;
#include <algorithm>
#include <string>
#include <iostream>
#include <cctype>
int main()
{
std::string str1 = "Text with some spaces";
std::string::iterator it = remove(str1.begin(), str1.end(), 't');
std::cout << str1 << std::endl;// prints "Tex wih some spaceses"
for (str1.begin();it != str1.end(); ++it)
{
std::cout << *it; //prints "es"
}
}
as you can see, the remove, only moves the lower case 't' to the end of the string, while returning a new iterator to the end of the new string (new string is the old string up to where the removed element are inserted)
this is why when you print the iterator that you got from "remove"
"Text with some spaces"
^ ^removes both 't', then shift all elements forward -1 //what we want to remove
"Text with some spaces"
^ end of string -2 //original state of string
"Tex with some spacess"
^end of string -3 //first 't' removed
"Tex wih some spaceses"
^end of string -4 //second 't' removed
"Tex wih some spaceses"
^new iterator that remove() returned -5 // the state of string after "remove" and without "erase"
if you pass the iterator you obtained from step 5 to "erase()" it will know to erase from there to the end of string re-sizing the string in process
To remove element with some condition(equal some value or other condition like less than) in container like vector, it always combine function member function erase and std::remove or std::remove_if.
In vector, the function erase can just delete element by position, like:
iterator erase (iterator position);
iterator erase (iterator first, iterator last);
But if you want to erase elements with some condition, you can combine it with std::remove or std::remove_if.
For example, you want to erase all the elements 6 in the below vector:
std::vector<int> vec{6, 8, 10, 3, 4, 5, 6, 6, 6, 7, 8};
// std::remove move elements and return iterator for vector erase funtion
auto last = std::remove(vec.begin(), vec.end(), 6);
for(int a:vec)
cout<<a<<" ";
cout<<endl;
// 8 10 3 4 5 7 8 6 6 7 8
vec.erase(last, vec.end());
for(int a:vec)
cout<<a<<" ";
cout<<endl;
// 8 10 3 4 5 7 8
std::remove works as below, it does't erase any elements, it just move elements and returns the iterator.
Possible implementation:
template< class ForwardIt, class T >
ForwardIt remove(ForwardIt first, ForwardIt last, const T& value)
{
first = std::find(first, last, value);
if (first != last)
for(ForwardIt i = first; ++i != last; )
if (!(*i == value))
*first++ = std::move(*i);
return first;
}
Conclusion:
If you want to remove elements with some condition, you use vector::iterator erase (iterator first, iterator last); essentially.
First get range start:
auto last = std::remove(vec.begin(), vec.end(), equal_condition_value);
erase by range(always with end())
vec.erase(last, vec.end());
cited:
https://en.cppreference.com/w/cpp/algorithm/remove
Simplest I can come up with:
erase() is something you can do to an element in a container. Given an iterator/index into a container, erase( it ) removes the thing the iterator refers to from the container.
remove() is something you can do to a range, it re-arranges that range but doesn't
erase anything from the range.
remove doesn't "really" remove
anything, because it can't.
In order to "actually" remove the elements from container you need to access container APIs. Where as remove works only with iterators irrespective of what containers those iterators points to. Hence, even if remove wants an "actual remove", it can't.
Remove overwrite "removed" elements by the following elements that were not removed and then it is up to the caller to decide to use the returned new logical end instead of the original end.
In your case remove logically removed 1 from vector a but size remained to 2 itself. Erase actually deleted the elements from vector. [ from vector new end to old end ]
The main idea of remove is it cannot change the number of elements and it just remove elements from a range as per criteria.