std::vector<int *>intList;
int *a = new int;
*a = 3;
int *b = new int;
*b = 2;
int *c = new int;
*c = 1;
intList.push_back(a);
intList.push_back(b);
intList.push_back(c);
std::vector<int *>::iterator it;
for (it = intList.begin(); it != intList.end();)
{
int *a = (int*)*it;
if ((*a) == 2)
{
delete a;
intList.erase(it);
}
else
{
++it;
}
}
This code crashes at the start of the for loop when something is erased from the vector. I do not understand why. I am using Visual Studio 2015 if that helps
erase returns next iterator.
if ((*a) == 2)
{
delete a;
it = intList.erase(it);
}
EDIT:
remove() and remove_if() will copy the elements(pointer here) and one will end up with multiple elements pointing to same integer and if you then try to free them, you'll be left with dangling pointers.
Consider the vector has 4 elements which look something like
0x196c160 0x196bec0 0x196c180 0x196bee0
One might be tempted to use erase-remove idiom
auto temp = remove_if(vec.begin(),vec.end(),[](const auto &i){return *i==2;});
Now it looks like
0x144aec0 0x144b180 0x144b180 0x144aee0
temp would be pointing to 3rd element and a
for(auto it=temp;it!=vec.end();it++)
delete *it;
Now the second element is a dangling pointer.
EDIT 2:
The above problem could be solved if you delete before the element is copied.Look at #Dietmar's answer.
Better to use std::vector<std::unique_ptr<int>> (or even std::vector<int> if you don't need pointer).
then just use erase-remove idiom:
std::vector<int> intList{3, 2, 1};
intList.erase(std::remove(intList.begin(), intList.end(), 2), intList.end());
or
std::vector<std::unique_ptr<int>> intList;
intList.puch_back(std::make_unique<int>(3));
intList.puch_back(std::make_unique<int>(2));
intList.puch_back(std::make_unique<int>(1));
intList.erase(std::remove_if(intList.begin(), intList.end(),
[](const auto& p){ return *p == 2; }),
intList.end());
If you really need raw owning pointer, you may use a variant using partition:
std::vector<int*> intList{ new int {3}, new int {2}, new int{1} };
auto it = std::partition(intList.begin(), intList.end(),
[](const auto& p){ return *p != 2; });
std::foreach (it, intList.end(), [](int* p) { delete p; });
intList.erase(it, intList.end());
Finally, if you really need to do it manually, you have to fix your erase line to:
it = intList.erase(it);
to have:
std::vector<int*> intList{ new int {3}, new int {2}, new int{1} };
for (auto it = intList.begin(); it != intList.end(); /*Empty*/) {
int *p = *it;
if (*p == 2) {
delete p;
it = intList.erase(it);
} else {
++it;
}
}
That code is causing iterator invalidation.
You're deleting a then expecting your iterator (which is just a pointer) to know what just happened.
If you have to iterate through the whole vector then consider a temporary vector of what to keep/throwaway and use that.
Or better yet just use find
http://www.cplusplus.com/reference/algorithm/find/
The simple anser is: you don’t! Instead you use a two stage approach: first get rid of the elements, then resize the container. The use of raw pointers slightly complicates things but it is still doable:
auto end = std::remove_if(intList.begin(), intList.end(),
[](int *ptr){ return *ptr == 2 && (delete ptr, true); });
intList.erase(end, endList.end());
Trying to erase individual elements while iterating over std::vector has non-linear worst case complexity.
If you take a look at documentation the erase function returns next iterator. In your case using it = intList.erase(it) is the solution. After c++11 all erase functions from other containers follow the same idea.
Related
std::vector<int *>intList;
int *a = new int;
*a = 3;
int *b = new int;
*b = 2;
int *c = new int;
*c = 1;
intList.push_back(a);
intList.push_back(b);
intList.push_back(c);
std::vector<int *>::iterator it;
for (it = intList.begin(); it != intList.end();)
{
int *a = (int*)*it;
if ((*a) == 2)
{
delete a;
intList.erase(it);
}
else
{
++it;
}
}
This code crashes at the start of the for loop when something is erased from the vector. I do not understand why. I am using Visual Studio 2015 if that helps
erase returns next iterator.
if ((*a) == 2)
{
delete a;
it = intList.erase(it);
}
EDIT:
remove() and remove_if() will copy the elements(pointer here) and one will end up with multiple elements pointing to same integer and if you then try to free them, you'll be left with dangling pointers.
Consider the vector has 4 elements which look something like
0x196c160 0x196bec0 0x196c180 0x196bee0
One might be tempted to use erase-remove idiom
auto temp = remove_if(vec.begin(),vec.end(),[](const auto &i){return *i==2;});
Now it looks like
0x144aec0 0x144b180 0x144b180 0x144aee0
temp would be pointing to 3rd element and a
for(auto it=temp;it!=vec.end();it++)
delete *it;
Now the second element is a dangling pointer.
EDIT 2:
The above problem could be solved if you delete before the element is copied.Look at #Dietmar's answer.
Better to use std::vector<std::unique_ptr<int>> (or even std::vector<int> if you don't need pointer).
then just use erase-remove idiom:
std::vector<int> intList{3, 2, 1};
intList.erase(std::remove(intList.begin(), intList.end(), 2), intList.end());
or
std::vector<std::unique_ptr<int>> intList;
intList.puch_back(std::make_unique<int>(3));
intList.puch_back(std::make_unique<int>(2));
intList.puch_back(std::make_unique<int>(1));
intList.erase(std::remove_if(intList.begin(), intList.end(),
[](const auto& p){ return *p == 2; }),
intList.end());
If you really need raw owning pointer, you may use a variant using partition:
std::vector<int*> intList{ new int {3}, new int {2}, new int{1} };
auto it = std::partition(intList.begin(), intList.end(),
[](const auto& p){ return *p != 2; });
std::foreach (it, intList.end(), [](int* p) { delete p; });
intList.erase(it, intList.end());
Finally, if you really need to do it manually, you have to fix your erase line to:
it = intList.erase(it);
to have:
std::vector<int*> intList{ new int {3}, new int {2}, new int{1} };
for (auto it = intList.begin(); it != intList.end(); /*Empty*/) {
int *p = *it;
if (*p == 2) {
delete p;
it = intList.erase(it);
} else {
++it;
}
}
That code is causing iterator invalidation.
You're deleting a then expecting your iterator (which is just a pointer) to know what just happened.
If you have to iterate through the whole vector then consider a temporary vector of what to keep/throwaway and use that.
Or better yet just use find
http://www.cplusplus.com/reference/algorithm/find/
The simple anser is: you don’t! Instead you use a two stage approach: first get rid of the elements, then resize the container. The use of raw pointers slightly complicates things but it is still doable:
auto end = std::remove_if(intList.begin(), intList.end(),
[](int *ptr){ return *ptr == 2 && (delete ptr, true); });
intList.erase(end, endList.end());
Trying to erase individual elements while iterating over std::vector has non-linear worst case complexity.
If you take a look at documentation the erase function returns next iterator. In your case using it = intList.erase(it) is the solution. After c++11 all erase functions from other containers follow the same idea.
I want to clear a element from a vector using the erase method. But the problem here is that the element is not guaranteed to occur only once in the vector. It may be present multiple times and I need to clear all of them. My code is something like this:
void erase(std::vector<int>& myNumbers_in, int number_in)
{
std::vector<int>::iterator iter = myNumbers_in.begin();
std::vector<int>::iterator endIter = myNumbers_in.end();
for(; iter != endIter; ++iter)
{
if(*iter == number_in)
{
myNumbers_in.erase(iter);
}
}
}
int main(int argc, char* argv[])
{
std::vector<int> myNmbers;
for(int i = 0; i < 2; ++i)
{
myNmbers.push_back(i);
myNmbers.push_back(i);
}
erase(myNmbers, 1);
return 0;
}
This code obviously crashes because I am changing the end of the vector while iterating through it. What is the best way to achieve this? I.e. is there any way to do this without iterating through the vector multiple times or creating one more copy of the vector?
Use the remove/erase idiom:
std::vector<int>& vec = myNumbers; // use shorter name
vec.erase(std::remove(vec.begin(), vec.end(), number_in), vec.end());
What happens is that remove compacts the elements that differ from the value to be removed (number_in) in the beginning of the vector and returns the iterator to the first element after that range. Then erase removes these elements (whose value is unspecified).
Edit: While updating a dead link I discovered that starting in C++20 there are freestanding std::erase and std::erase_if functions that work on containers and simplify things considerably.
Calling erase will invalidate iterators, you could use:
void erase(std::vector<int>& myNumbers_in, int number_in)
{
std::vector<int>::iterator iter = myNumbers_in.begin();
while (iter != myNumbers_in.end())
{
if (*iter == number_in)
{
iter = myNumbers_in.erase(iter);
}
else
{
++iter;
}
}
}
Or you could use std::remove_if together with a functor and std::vector::erase:
struct Eraser
{
Eraser(int number_in) : number_in(number_in) {}
int number_in;
bool operator()(int i) const
{
return i == number_in;
}
};
std::vector<int> myNumbers;
myNumbers.erase(std::remove_if(myNumbers.begin(), myNumbers.end(), Eraser(number_in)), myNumbers.end());
Instead of writing your own functor in this case you could use std::remove:
std::vector<int> myNumbers;
myNumbers.erase(std::remove(myNumbers.begin(), myNumbers.end(), number_in), myNumbers.end());
In C++11 you could use a lambda instead of a functor:
std::vector<int> myNumbers;
myNumbers.erase(std::remove_if(myNumbers.begin(), myNumbers.end(), [number_in](int number){ return number == number_in; }), myNumbers.end());
In C++17 std::experimental::erase and std::experimental::erase_if are also available, in C++20 these are (finally) renamed to std::erase and std::erase_if (note: in Visual Studio 2019 you'll need to change your C++ language version to the latest experimental version for support):
std::vector<int> myNumbers;
std::erase_if(myNumbers, Eraser(number_in)); // or use lambda
or:
std::vector<int> myNumbers;
std::erase(myNumbers, number_in);
You can iterate using the index access,
To avoid O(n^2) complexity
you can use two indices, i - current testing index, j - index to
store next item and at the end of the cycle new size of the vector.
code:
void erase(std::vector<int>& v, int num)
{
size_t j = 0;
for (size_t i = 0; i < v.size(); ++i) {
if (v[i] != num) v[j++] = v[i];
}
// trim vector to new size
v.resize(j);
}
In such case you have no invalidating of iterators, complexity is O(n), and code is very concise and you don't need to write some helper classes, although in some case using helper classes can benefit in more flexible code.
This code does not use erase method, but solves your task.
Using pure stl you can do this in the following way (this is similar to the Motti's answer):
#include <algorithm>
void erase(std::vector<int>& v, int num) {
vector<int>::iterator it = remove(v.begin(), v.end(), num);
v.erase(it, v.end());
}
Depending on why you are doing this, using a std::set might be a better idea than std::vector.
It allows each element to occur only once. If you add it multiple times, there will only be one instance to erase anyway. This will make the erase operation trivial.
The erase operation will also have lower time complexity than on the vector, however, adding elements is slower on the set so it might not be much of an advantage.
This of course won't work if you are interested in how many times an element has been added to your vector or the order the elements were added.
There are std::erase and std::erase_if since C++20 which combines the remove-erase idiom.
std::vector<int> nums;
...
std::erase(nums, targetNumber);
or
std::vector<int> nums;
...
std::erase_if(nums, [](int x) { return x % 2 == 0; });
If you change your code as follows, you can do stable deletion.
void atest(vector<int>& container,int number_in){
for (auto it = container.begin(); it != container.end();) {
if (*it == number_in) {
it = container.erase(it);
} else {
++it;
}
}
}
However, a method such as the following can also be used.
void btest(vector<int>& container,int number_in){
container.erase(std::remove(container.begin(), container.end(), number_in),container.end());
}
If we must preserve our sequence’s order (say, if we’re keeping it sorted by some interesting property), then we can use one of the above. But if the sequence is just a bag of values whose order we don’t care about at all, then we might consider moving single elements from the end of the sequence to fill each new gap as it’s created:
void ctest(vector<int>& container,int number_in){
for (auto it = container.begin(); it != container.end(); ) {
if (*it == number_in) {
*it = std::move(container.back());
container.pop_back();
} else {
++it;
}
}
}
Below are their benchmark results:
CLang 15.0:
Gcc 12.2:
I am new in programming c++, so please don't be angry with me if my source code is not exactly brilliant.
I have to write a programm to handle with nodes and edges in a graph for my studies.
I have 2 std::lists in my source code. The first one is to store general Nodes and the other one for saving the kind class of my nodes called ArticleNodes. In general, all elements are pointers to the created objects.
To figure out whether one object is the same in the other list I save the memory address and compare it to the elements on the second list. If there is a match the second element will be deleted.
Now I'd like to delete one element in both lists:
void Graph::deleteNode(unsigned int nodeNumber)
{
list<Node*>::iterator it = m_nodes.begin();
ArticleNode* pCurrentArticleNode;
for(unsigned int i=1; i<nodeNumber; i++) { it++; }
Node* pCurrentNode = (*it);
for (list<ArticleNode*>::iterator itArticle = m_articlenode.begin(); itArticle != m_articlenode.end(); itArticle++)
{
pCurrentArticleNode = (*itArticle);
if(pCurrentNode==pCurrentArticleNode) { m_articlenode.remove(pCurrentArticleNode); }
}
m_nodes.remove(pCurrentNode);
delete pCurrentNode;
delete pCurrentArticleNode;
}
I can compile this, but when I call the function, my programm just exits with return 1.
Actually, I figured out that the remove-command in the if-clause is the problem. Why does that not work??
You should use algorithms more than doing everything manually:
void Graph::deleteNode(unsigned int nodeNumber)
{
assert (nodeNumber < m_nodes.size());
auto it = std::next( m_nodes.begin(), nodeNumber - 1 );
auto itArticle = std::find( m_articlenode.begin(), m_articlenode.end(), *it );
if( itArticle != m_articlenode.end() )
m_articlenode.erase( itArticle );
delete *it;
m_nodes.erase(it);
}
Btw your code deletes the same object twice.
When you remove an element from std::list object with remove( ) method, all iterators pointing to that elements become invalid. In your case, after you remove an element from the list m_articlenode, the iterator object itArticle becomes invalid. And when you increment that iterator, you get an undefined behavior.
Pay attention that the method remove( ) deletes all of the items in the list with given value. So you don't need the for-loop at all. Here is the fixed version of your function:
void Graph::deleteNode(unsigned int nodeNumber)
{
list<Node*>::iterator it = m_nodes.begin();
for(unsigned int i=1; i<nodeNumber; i++) { it++; }
Node* pCurrentNode = (*it);
m_articlenode.remove(pCurrentNode);
m_nodes.remove(pCurrentNode);
delete pCurrentNode;
}
You have to use the erase method of std::list in order to remove an element from your list while iterating over it.
This should do the trick:
void Graph::deleteNode(unsigned int nodeNumber)
{
list<Node*>::iterator it = m_nodes.begin();
ArticleNode* pCurrentArticleNode;
for(unsigned int i=1; i<nodeNumber; i++) { it++; }
Node* pCurrentNode = (*it);
list<ArticleNode*>::iterator itArticle = m_articlenode.begin();
while(itArticle != m_articlenode.end()) {
pCurrentArticleNode = (*itArticle);
if(pCurrentNode==pCurrentArticleNode) {
m_articlenode.erase(itArticle++);
} else {
itArticle++;
}
}
m_nodes.remove(pCurrentNode);
delete pCurrentNode;
delete pCurrentArticleNode;
}
Simple issue in your code is that if you go into the if condition is met and it's body executed, you should come out of the loop. std::remove invalidates the iterator ann you will get issues in next iteration so do:
for (list<ArticleNode*>::iterator itArticle = m_articlenode.begin(); itArticle != m_articlenode.end(); itArticle++)
{
pCurrentArticleNode = (*itArticle);
if(pCurrentNode==pCurrentArticleNode)
{
m_articlenode.remove(pCurrentArticleNode);
break;
}
}
In general, there are other issues in the code. As a first step I would suggest using shared_ptr for managing your Nodes and have list of shared_ptr instead of list of pointers.
I am new in programming c++, so please don't be angry with me if my
source code is not exactly brilliant.
We all start somewhere.
Now I'd like to delete one element in both lists:
OK. Quick question. Why are you deleting article node if it is already deleted (via base Node)? I'm assuming for now node is not duplicated in the list:
My solution below... I've passed the lists as arguments. See comments:
#include <list>
#include <algorithm>
struct Node
{
virtual ~Node(){} //For dyna cast to work...
};
struct ArticleNode : Node
{
};
void deleteNode(std::list<ArticleNode*>& articleList, std::list<Node*>& m_nodes, unsigned int nodeNumber)
{
using namespace std;
if (m_nodes.size() > nodeNumber)
{
auto it = m_nodes.begin();
// Advance advances our iterator by N. No need for your for loop - less risk...
std::advance(it,nodeNumber);
Node* currentNode = *it;
//Casting is bad here, but hey, lets assume if type is wrong, we only erase
// it from Node...(Your call)?
ArticleNode* currentArticleNode = dynamic_cast<ArticleNode*>(currentNode);
if (currentArticleNode)
{
//Use find here.... KISS
auto foundPos = std::find(articleList.begin(), articleList.end(), currentArticleNode);
if (foundPos != articleList.end())
{
//No need to delete currentArticleNode, as we're deleting it already...
articleList.erase(foundPos);
}
//Assuming only one item for now...
}
//Else our node was obviously not the right type, and cannot exist in articleNodes...
m_nodes.erase(it);
delete currentNode;
}
else
{
std::cout << "No such node: " << nodeNumber << std::endl;
}
}
I am wondering what the fast swap method is in C++ containers such as list and vector, because I haven't found any builtin swap functions yet.FYI, I want to swap object rather than the whole list.
For example, assume we have such int sequence 3 2 4 5 and they are stored in a list container(stl), and I want to swap 2 and 4. Here is the dumb method that I came up with:
list<int> numbers;
numbers.push_back(3);
numbers.push_back(2);
numbers.push_back(4);
numbers.push_back(5);
list<int>::iterator item;
item=numbers.begin();
advance(item,2);
int key = *item;
advance(item,-1);
numbers.insert(item,key);
advance(item,1);
numbers.erase(item);
So, briefly speaking, what I am doing here is just "Copy, insert, and remove", and the reason why I do this is that I heard list container is very efficient for inserting and removing elements,but I am pretty sure there should be better algorithms.In addition, I also heard there exists a constant time swap method related to pointers, so anyone knows anything about it?
Thanks for helping out.
You want std::swap:
list<int>::iterator item1 = numbers.begin();
++item1;
list<int>::iterator item2 = item1;
++item2;
std::swap(*item1, *item2);
Use iter_swap to swap the elements pointed at by two iterators to the list. This swaps the data, rather than the nodes, but it's easy.
#include <list>
#include <iostream>
int main() {
std::list<int> numbers;
numbers.push_back(3);
numbers.push_back(2);
numbers.push_back(4);
numbers.push_back(5);
auto first = std::next(numbers.begin(), 2);
auto second = std::next(numbers.begin(), 1);
std::iter_swap(first, second);
for(int& v : numbers)
std::cout << v << ' ';
}
http://coliru.stacked-crooked.com/view?id=a89b3b1ae9400367b6ff194d1b504e58-f674c1a6d04c632b71a62362c0ccfc51
If you want to swap nodes rather than elements, you can use list::splice, though it's a little tricker:
int main() {
std::list<int> numbers;
numbers.push_back(3);
numbers.push_back(2);
numbers.push_back(4);
numbers.push_back(5);
std::list<int> temporary;
auto move_from = std::next(numbers.begin(), 2);
temporary.splice(temporary.begin(), numbers, move_from, std::next(move_from));
auto move_to = std::next(numbers.begin(), 1);
numbers.splice(move_to, temporary);
for(int& v : numbers)
std::cout << v << ' ';
}
It seems you may be looking for a way to move around nodes within the list, without copying actual elements. You could do this with list::splice. Nothing like that is possible for vector, of course, which is not node-based.
Something like this:
list<int>::iterator to = numbers.begin();
++to;
list<int>::iterator which = to;
++which;
numbers.splice(to, numbers, which);
How about using swap?
using std::swap;
swap(numbers[1], numbers[2]);
which will use std:swap or ADL to determine a proper swap-function if there is one defined for the arguments.
As #Mooing Duck correctly pointed out std::list requires you to use iterators.
std::iter_swap(numbers.begin()+1, numbers.begin()+2);
You can also use
using std::swap;
std::list<int>::iterator item(numbers.begin());
std::advance(item, 1);
std::list<int>::iterator other(item);
std::advance(other, 1);
swap(*item, *other);
or
using std::swap;
swap(*std::next(numbers.begin(), 1), *std::next(numbers.begin(), 2));
or
std::iter_swap(std::next(numbers.begin(), 1), std::next(numbers.begin(), 2));
With std::swap(),
int reverse(std::list<int>& list){
int exchange = 0;
int move = 0;
int distance_lo = 0;
int distance_hi = 0;
std::list<int>::iterator it_lo = list.begin();
std::list<int>::iterator it_hi = --list.end();
while (1) {
it_lo = list.begin();
it_hi = --list.end();
std::advance(it_lo, move);
std::advance(it_hi, -move);
distance_lo = std::distance(list.begin(), it_lo);
distance_hi = std::distance(list.begin(), it_hi);
if (distance_lo < distance_hi) {
std::swap(*it_lo, *it_hi);
exchange++;
} else {
break;
}
move++;
}
return exchange; }
With std::list::splice(),
int reverse(std::list<int>& list) {
int exchange = 0;
int move = 0;
int distance_lo = 0;
int distance_hi = 0;
std::list<int>::iterator it_lo = list.begin();
std::list<int>::iterator it_hi = --list.end();
while (1) {
it_lo = list.begin();
it_hi = --list.end();
std::advance(it_lo, move);
std::advance(it_hi, -move);
distance_lo = std::distance(list.begin(), it_lo);
distance_hi = std::distance(list.begin(), it_hi);
if (distance_lo < distance_hi) {
std::list<int> tmp;
tmp.splice(tmp.begin(), list, it_lo);
tmp.splice(std::next(tmp.begin(),1), list, it_hi);
it_lo = list.begin();
it_hi = --list.end();
std::advance(it_lo, move); //insert before it_lo
std::advance(it_hi, -move);
std::advance(it_hi, 1); //insert after it_hi
list.splice(it_lo, tmp, std::next(tmp.begin(),1));
list.splice(it_hi, tmp, tmp.begin());
exchange++;
} else {
break;
}
move++;
}
return exchange; }
Hope it could help you :)
I have this loop
for(int i=0;i<vec1.size();++i)
{
if(vec1[i]==*p)
{
vec1[i]=*p;
cout<<"element updated"<<endl;
}
else
{
cout<<"push_back"<<endl;
vec1.push_back(*p);
}
}
I'm inserting objects in container class and I've overloaded the == to check two parameters inside the object and if they match I want to update the them and if they don't match I want to put them in the vector, but I don't seem to be able to properly populate my vector, when I do vec1.size() I get 0 even when I insert 3 objects.
You're problem is that your if is inside your search loop. Your if will never be executed, because your loop body never runs, because your .size() will never be greater than 0.
Try this:
// UNTESTED
std::vector<person> vec1;
add(person *p) {
std::vector<person>::iterator it = std::find(vec1.begin(), vec1.end(), *p);
if(it == vec1.end())
vec1.push_back(*p);
else
*it = *p;
}
Or, if you really want to code the loop by hand:
// UNTESTED
std::vector<person> vec1;
add(person *p) {
int i;
for(i=0;i<vec1.size();++i) {
if(vec1[i] == *p)
break;
}
if(i == vec1.size())
vec1.push_back(*p);
else
vec1[i] = *p;
}
Of course, you might consider changing your container. Using a std::map would shorten your code and reduce the time it takes to manipulate large data sets.
std::map<std::string, person> map1;
add(person *p) {
map1[p->name] = *p;
}
When the vec1 starts from empty, the for loop is not going to run. So you want to have at least one element in vec1 to start with. How about add this:
vec1.push_back(*p);
for(int i=0;i<vec1.size();++i){//the rest}