I've got code that looks like this:
for (std::list<item*>::iterator i=items.begin();i!=items.end();i++)
{
bool isActive = (*i)->update();
//if (!isActive)
// items.remove(*i);
//else
other_code_involving(*i);
}
items.remove_if(CheckItemNotActive);
I'd like remove inactive items immediately after update them, inorder to avoid walking the list again. But if I add the commented-out lines, I get an error when I get to i++: "List iterator not incrementable". I tried some alternates which didn't increment in the for statement, but I couldn't get anything to work.
What's the best way to remove items as you are walking a std::list?
You have to increment the iterator first (with i++) and then remove the previous element (e.g., by using the returned value from i++). You can change the code to a while loop like so:
std::list<item*>::iterator i = items.begin();
while (i != items.end())
{
bool isActive = (*i)->update();
if (!isActive)
{
items.erase(i++); // alternatively, i = items.erase(i);
}
else
{
other_code_involving(*i);
++i;
}
}
You want to do:
i= items.erase(i);
That will correctly update the iterator to point to the location after the iterator you removed.
You need to do the combination of Kristo's answer and MSN's:
// Note: Using the pre-increment operator is preferred for iterators because
// there can be a performance gain.
//
// Note: As long as you are iterating from beginning to end, without inserting
// along the way you can safely save end once; otherwise get it at the
// top of each loop.
std::list< item * >::iterator iter = items.begin();
std::list< item * >::iterator end = items.end();
while (iter != end)
{
item * pItem = *iter;
if (pItem->update() == true)
{
other_code_involving(pItem);
++iter;
}
else
{
// BTW, who is deleting pItem, a.k.a. (*iter)?
iter = items.erase(iter);
}
}
Of course, the most efficient and SuperCool® STL savy thing would be something like this:
// This implementation of update executes other_code_involving(Item *) if
// this instance needs updating.
//
// This method returns true if this still needs future updates.
//
bool Item::update(void)
{
if (m_needsUpdates == true)
{
m_needsUpdates = other_code_involving(this);
}
return (m_needsUpdates);
}
// This call does everything the previous loop did!!! (Including the fact
// that it isn't deleting the items that are erased!)
items.remove_if(std::not1(std::mem_fun(&Item::update)));
I have sumup it, here is the three method with example:
1. using while loop
list<int> lst{4, 1, 2, 3, 5};
auto it = lst.begin();
while (it != lst.end()){
if((*it % 2) == 1){
it = lst.erase(it);// erase and go to next
} else{
++it; // go to next
}
}
for(auto it:lst)cout<<it<<" ";
cout<<endl; //4 2
2. using remove_if member funtion in list:
list<int> lst{4, 1, 2, 3, 5};
lst.remove_if([](int a){return a % 2 == 1;});
for(auto it:lst)cout<<it<<" ";
cout<<endl; //4 2
3. using std::remove_if funtion combining with erase member function:
list<int> lst{4, 1, 2, 3, 5};
lst.erase(std::remove_if(lst.begin(), lst.end(), [](int a){
return a % 2 == 1;
}), lst.end());
for(auto it:lst)cout<<it<<" ";
cout<<endl; //4 2
4. using for loop , should note update the iterator:
list<int> lst{4, 1, 2, 3, 5};
for(auto it = lst.begin(); it != lst.end();++it){
if ((*it % 2) == 1){
it = lst.erase(it); erase and go to next(erase will return the next iterator)
--it; // as it will be add again in for, so we go back one step
}
}
for(auto it:lst)cout<<it<<" ";
cout<<endl; //4 2
Use std::remove_if algorithm.
Edit:
Work with collections should be like:
prepare collection.
process collection.
Life will be easier if you won't mix this steps.
std::remove_if. or list::remove_if ( if you know that you work with list and not with the TCollection )
std::for_each
The alternative for loop version to Kristo's answer.
You lose some efficiency, you go backwards and then forward again when deleting but in exchange for the extra iterator increment you can have the iterator declared in the loop scope and the code looking a bit cleaner. What to choose depends on priorities of the moment.
The answer was totally out of time, I know...
typedef std::list<item*>::iterator item_iterator;
for(item_iterator i = items.begin(); i != items.end(); ++i)
{
bool isActive = (*i)->update();
if (!isActive)
{
items.erase(i--);
}
else
{
other_code_involving(*i);
}
}
Here's an example using a for loop that iterates the list and increments or revalidates the iterator in the event of an item being removed during traversal of the list.
for(auto i = items.begin(); i != items.end();)
{
if(bool isActive = (*i)->update())
{
other_code_involving(*i);
++i;
}
else
{
i = items.erase(i);
}
}
items.remove_if(CheckItemNotActive);
Removal invalidates only the iterators that point to the elements that are removed.
So in this case after removing *i , i is invalidated and you cannot do increment on it.
What you can do is first save the iterator of element that is to be removed , then increment the iterator and then remove the saved one.
If you think of the std::list like a queue, then you can dequeue and enqueue all the items that you want to keep, but only dequeue (and not enqueue) the item you want to remove. Here's an example where I want to remove 5 from a list containing the numbers 1-10...
std::list<int> myList;
int size = myList.size(); // The size needs to be saved to iterate through the whole thing
for (int i = 0; i < size; ++i)
{
int val = myList.back()
myList.pop_back() // dequeue
if (val != 5)
{
myList.push_front(val) // enqueue if not 5
}
}
myList will now only have numbers 1-4 and 6-10.
Iterating backwards avoids the effect of erasing an element on the remaining elements to be traversed:
typedef list<item*> list_t;
for ( list_t::iterator it = items.end() ; it != items.begin() ; ) {
--it;
bool remove = <determine whether to remove>
if ( remove ) {
items.erase( it );
}
}
PS: see this, e.g., regarding backward iteration.
PS2: I did not thoroughly tested if it handles well erasing elements at the ends.
You can write
std::list<item*>::iterator i = items.begin();
while (i != items.end())
{
bool isActive = (*i)->update();
if (!isActive) {
i = items.erase(i);
} else {
other_code_involving(*i);
i++;
}
}
You can write equivalent code with std::list::remove_if, which is less verbose and more explicit
items.remove_if([] (item*i) {
bool isActive = (*i)->update();
if (!isActive)
return true;
other_code_involving(*i);
return false;
});
The std::vector::erase std::remove_if idiom should be used when items is a vector instead of a list to keep compexity at O(n) - or in case you write generic code and items might be a container with no effective way to erase single items (like a vector)
items.erase(std::remove_if(begin(items), end(items), [] (item*i) {
bool isActive = (*i)->update();
if (!isActive)
return true;
other_code_involving(*i);
return false;
}));
do while loop, it's flexable and fast and easy to read and write.
auto textRegion = m_pdfTextRegions.begin();
while(textRegion != m_pdfTextRegions.end())
{
if ((*textRegion)->glyphs.empty())
{
m_pdfTextRegions.erase(textRegion);
textRegion = m_pdfTextRegions.begin();
}
else
textRegion++;
}
I'd like to share my method. This method also allows the insertion of the element to the back of the list during iteration
#include <iostream>
#include <list>
int main(int argc, char **argv) {
std::list<int> d;
for (int i = 0; i < 12; ++i) {
d.push_back(i);
}
auto it = d.begin();
int nelem = d.size(); // number of current elements
for (int ielem = 0; ielem < nelem; ++ielem) {
auto &i = *it;
if (i % 2 == 0) {
it = d.erase(it);
} else {
if (i % 3 == 0) {
d.push_back(3*i);
}
++it;
}
}
for (auto i : d) {
std::cout << i << ", ";
}
std::cout << std::endl;
// result should be: 1, 3, 5, 7, 9, 11, 9, 27,
return 0;
}
I think you have a bug there, I code this way:
for (std::list<CAudioChannel *>::iterator itAudioChannel = audioChannels.begin();
itAudioChannel != audioChannels.end(); )
{
CAudioChannel *audioChannel = *itAudioChannel;
std::list<CAudioChannel *>::iterator itCurrentAudioChannel = itAudioChannel;
itAudioChannel++;
if (audioChannel->destroyMe)
{
audioChannels.erase(itCurrentAudioChannel);
delete audioChannel;
continue;
}
audioChannel->Mix(outBuffer, numSamples);
}
public static void get_sum_while (int[] num,int len)
{
int sum2,i=0;
while ( i<len)
{
sum2=sum2+num[i];
i++;
}
Console.WriteLine("The sum of the series by while loop is {0}",sum2);
}
public static int get_sum_recur (int[] num,int len)
{
int sum3;
if (len==0)
return sum3=sum3+num[0];
else
{
return sum3=num[len]+get_sum_recur(num,length-1);
}
}
}
Hello this gives sum of the series from three function the first two are okay but recursive did not give it give exception i don't where i go wrong and is it correct way to get sum by recursion?
The idea of summing a list by recursion is to sum the an element of the list with the sum of the same list without the chosen element.
sum([a,b,c,d,e]) = a + sum([b,c,d,e])
So the initial value of the result have to be set to 0.
Then choose an element, for example the first one, add it to the current result and call sum on the rest of the list with the new result.
When list is empty, end recursion.
In pseudo code, because I don't have a C# compiler, this gives :
public static int get_sum_recur(int result, int[] list) {
if (len(list)==0) {
return result; // end recursion
}
else {
return get_sum_recur(result+list[0], list[1:]);
}
}
public static main {
print(get_sum_recur(0, [1,2,3,4,5,6])
}
My code is supposed to simultanously start sorting 3 different lists using different methods and return the first one to finish. However it always performs the first task on the list instead. How can I fix that?
Below is part of my code which seemed relevant to show.
static List<Task<List<int>>> listoftasks = new List<Task<List<int>>>() { QuickSortAsync(list1), BubbleSortAsync(list2), SelectionSortAsync(list3) };
public async static void caller()
{
List<int> result = await Task.WhenAny(listoftasks).Result;
foreach (var item in result)
Console.Write(item + ", ");
}
static Task<List<int>> QuickSortAsync(List<int> l)
{
return Task.Run<List<int>>(() =>
{
l.Sort();
return l;
});
}
Since your list of tasks is static, you're starting all three tasks very early. Then, when you call WhenAny, it's likely that they've already all completed.
I suggest you start the tasks when you call WhenAny:
public static async Task CallerAsync()
{
List<int> result = await await Task.WhenAny(QuickSortAsync(list1),
BubbleSortAsync(list2), SelectionSortAsync(list3));
foreach (var item in result)
Console.Write(item + ", ");
}
I have a problem that I'm importing some data and each new row depends on previous row being added (since each row has its order attribute set based on current maximum order from other objects). The flow is that I first try to find object with the same name, if not found I first check maximum order and create new object with order + 1 from that query.
I tried doing this with Q promises which are available under Waterline. I tried using all method as well as combining queries with then from Q docs:
var result = Q(initialVal);
funcs.forEach(function (f) {
result = result.then(f);
});
return result;
But all objects had the same order, just like they would be executed in parallel, instead of waiting for the first chain to finish.
I finally found a solution with recurrency, but I doubt it's the best way of working with promises. Here's the code that works (+ needs some refactor and cleaning etc.), to show the rough idea:
function findOrCreateGroup(groupsQuery, index, callback) {
var groupName = groupsQuery[index];
Group.findOne({ 'name' : groupName }).then(function(group) {
if (!group) {
return Group.find().limit(1).sort('order DESC').then(function(foundGroups) {
var maxOrder = 0;
if (foundGroups.length > 0) {
maxOrder = foundGroups[0].order;
}
return Group.create({
'name' : groupName,
'order' : (maxOrder + 1)
}).then(function(g) {
dbGroups[g.name] = g;
if (index + 1 < groupsQuery.length) {
findOrCreateGroup(groupsQuery, index + 1, callback);
} else {
callback();
}
return g;
});
});
} else {
dbGroups[group.name] = group;
if (index + 1 < groupsQuery.length) {
findOrCreateGroup(groupsQuery, index + 1, callback);
} else {
callback();
}
return group;
}
});
}
I'd like to know if two lists share values before applying an intersection. Something like bool DoIntersect(listA, listB) would be fabulous!
This is the code I came up with:
// Person is a class with Id and Name properties
List<Person> people1;
List<Person> people2;
// Populate people1 and people2...
// My current solution (pseudocode obviously)...
if (DoIntersect(people1, people2))
{
people1 = people1.Intersect(people2)
}
else
{
/* No shared people */
throw exception;
}
// Continue with the process...
It depends on exactly what you want:
// are there any common values between a and b?
public static bool SharesAnyValueWith<T>(this IEnumerable<T> a, IEnumerable<T> b)
{
return a.Intersect(b).Any();
}
For lists that don't overlap, this will iterate through a and b each once. For lists that overlap, this will iterate all the way through a, then through b until the first overlapping element is found.
// does a contain all of b? (ignores duplicates)
public static bool ContainsAllFrom<T>(this IEnumerable<T> a, IEnumerable<T> b)
{
return !b.Except(a).Any();
}
This will iterate through a once, then will iterate through b, stopping on the first element in b not in a.
// does a contain all of b? (considers duplicates)
public static bool ContainsAllFrom<T>(this IEnumerable<T> a, IEnumerable<T> b)
{
// get the count of each distinct element in a
var counts = a.GroupBy(t => t).ToDictionary(g => g.Key, g => g.Count());
foreach (var t in b) {
int count;
// if t isn't in a or has too few occurrences return false. Otherwise, reduce
// the count by 1
if (!counts.TryGetValue(t, out count) || count == 0) { return false; }
counts[t] = count - 1;
}
return true;
}
Similarly, this will iterate through a once, then will iterate through b, stopping on the first element in b not in a.
I believe without altering the fact that you're using a List you can't get better performance.
However, if you would have 2 sorted lists to begin with (requires overhead when creating them), then you could iterate through them with complexity of O(n) in order to find out if you have shared values.
Edit:
Although original OP doesn't have 2 sorted lists, in case someone will need it, here is the implementation for checking Intersection at O(n):
public Boolean DoIntersect(SortedList<int,String> listA,SortedList<int,String> listB )
{
if (listA == null || listA.Count == 0 || listB == null || listB.Count == 0)
{
return false;
}
var keysA = listA.Keys;
var keysB = listB.Keys;
int i = 0, j = 0;
while (i < listA.Count && j < listB.Count)
{
if (keysA[i] < keysB[j])
{
i++;
}else if (keysA[i] > keysB[j])
{
j++;
}
else
{
return true;
}
}
The above approach can be used also with IEnumerable lists, given that they are sorted, with slight variation - using GetEnumerator and iterating with it.