I'm stumped on my loop. When two lists sizes are equal, I want to compare the contents of it (string and int). I primarily don't understand this part:
The content of all the containers in a BookList is the same - so pick one to walk. If the books are different, you have your answer
Here is my code:
int BookList::compare(const BookList& other) const {
if (!containers_are_consistent() || !other.containers_are_consistent()) {
throw BookList::InvalidInternalStateException(
"Container consistency error in compare");
}
// my implementation starts here
auto begin_this_ = this->books_vector_.begin();
auto begin_other_ = other.books_vector_.begin();
auto end_this_ = this->books_vector_.end();
auto end_other_ = other.books_vector_.end();
if(this->size() == other.size()){
while(begin_this_ != end_this_) {
if(begin_this_ == begin_other_){
++begin_this_;
}
return 0;
if(begin_this_ != begin_other_) {
//what do I do here?
}
}
return 0;
} else if(this->size() < other.size()){
return -1;
} else if(this->size() > other.size()){
return 1;
}
// ends here
}
Firstly, you probably want to compare the contents of the iterators and not the iterators themselves
if (*begin_this_ == *begin_other_) ...
Secondly, you return 0 whenever two iterators compare equal which means you exit the loop.
I suggest you return early only if two elements are NOT equal.
Sadly, you have not described what value is returned if the sizes are equal but the contents are not, so I will assume that the elements are less than comparable.
Then your while loop will look like
while (begin_this_ != end_this_) {
if (*begin_this_ < *begin_other_)
return -1;
else if (*begin_other_ < *begin_this_)
return 1;
++begin_this_;
++begin_other_;
}
// end of loop which means that all elements are equal
return 0;
This function only removes repeated values if they are neighbour. How can it remove repeated values from all values and not just the neighbour value?
multimap<string, DictionaryItem> getDistinctI(multimap<string, DictionaryItem> distinct) {
if (distinct.size() > 1) {
auto it = begin(distinct);
auto previousIt = it;
++it;
while (it != end(distinct)) {
if (previousIt->second.getDefinition() == it->second.getDefinition()
&& previousIt->second.getPartOfSpeech() == it->second.getPartOfSpeech()) {
it = distinct.erase(it);
}
else {
previousIt = it;
++it;
}
}
}
return distinct;
}
This is a2.hpp, and is the program that can be edited, as far as I know the code is correct, just too slow. I am honestly lost here, I know my for loops are probably whats slowing me down so much, maybe use an iterator?
// <algorithm>, <list>, <vector>
// YOU CAN CHANGE/EDIT ANY CODE IN THIS FILE AS LONG AS SEMANTICS IS UNCHANGED
#include <algorithm>
#include <list>
#include <vector>
class key_value_sequences {
private:
std::list<std::vector<int>> seq;
std::vector<std::vector<int>> keyref;
public:
// YOU SHOULD USE C++ CONTAINERS TO AVOID RAW POINTERS
// IF YOU DECIDE TO USE POINTERS, MAKE SURE THAT YOU MANAGE MEMORY PROPERLY
// IMPLEMENT ME: SHOULD RETURN SIZE OF A SEQUENCE FOR GIVEN KEY
// IF NO SEQUENCE EXISTS FOR A GIVEN KEY RETURN 0
int size(int key) const;
// IMPLEMENT ME: SHOULD RETURN POINTER TO A SEQUENCE FOR GIVEN KEY
// IF NO SEQUENCE EXISTS FOR A GIVEN KEY RETURN nullptr
const int* data(int key) const;
// IMPLEMENT ME: INSERT VALUE INTO A SEQUENCE IDENTIFIED BY GIVEN KEY
void insert(int key, int value);
}; // class key_value_sequences
int key_value_sequences::size(int key) const {
//checks if the key is invalid or the count vector is empty.
if(key<0 || keyref[key].empty()) return 0;
// sub tract 1 because the first element is the key to access the count
return keyref[key].size() -1;
}
const int* key_value_sequences::data(int key) const {
//checks if key index or ref vector is invalid
if(key<0 || keyref.size() < static_cast<unsigned int>(key+1)) {
return nullptr;
}
// ->at(1) accesses the count (skipping the key) with a pointer
return &keyref[key].at(1);
}
void key_value_sequences::insert(int key, int value) {
//checks if key is valid and if the count vector needs to be resized
if(key>=0 && keyref.size() < static_cast<unsigned int>(key+1)) {
keyref.resize(key+1);
std::vector<int> val;
seq.push_back(val);
seq.back().push_back(key);
seq.back().push_back(value);
keyref[key] = seq.back();
}
//the index is already valid
else if(key >=0) keyref[key].push_back(value);
}
#endif // A2_HPP
This is a2.cpp, this just tests the functionality of a2.hpp, this code cannot be changed
// DO NOT EDIT THIS FILE !!!
// YOUR CODE MUST BE CONTAINED IN a2.hpp ONLY
#include <iostream>
#include "a2.hpp"
int main(int argc, char* argv[]) {
key_value_sequences A;
{
key_value_sequences T;
// k will be our key
for (int k = 0; k < 10; ++k) { //the actual tests will have way more than 10 sequences.
// v is our value
// here we are creating 10 sequences:
// key = 0, sequence = (0)
// key = 1, sequence = (0 1)
// key = 2, sequence = (0 1 2)
// ...
// key = 9, sequence = (0 1 2 3 4 5 6 7 8 9)
for (int v = 0; v < k + 1; ++v) T.insert(k, v);
}
T = T;
key_value_sequences V = T;
A = V;
}
std::vector<int> ref;
if (A.size(-1) != 0) {
std::cout << "fail" << std::endl;
return -1;
}
for (int k = 0; k < 10; ++k) {
if (A.size(k) != k + 1) {
std::cout << "fail";
return -1;
} else {
ref.clear();
for (int v = 0; v < k + 1; ++v) ref.push_back(v);
if (!std::equal(ref.begin(), ref.end(), A.data(k))) {
std::cout << "fail 3 " << A.data(k) << " " << ref[k];
return -1;
}
}
}
std::cout << "pass" << std::endl;
return 0;
} // main
If anyone could help me improve my codes efficiency I would really appreciate it, thanks.
First, I'm not convinced your code is correct. In insert, if they key is valid you create a new vector and insert it into sequence. Sounds wrong, as that should only happen if you have a new key, but if your tests pass it might be fine.
Performance wise:
Avoid std::list. Linked lists have terrible performance on today's hardware because they break pipelineing, caching and pre-fetching. Always use std::vector instead. If the payload is really big and you are worried about copies use std::vector<std::unique_ptr<T>>
Try to avoid copying vectors. In your code you have keyref[key] = seq.back() which copies the vector, but should be fine since it's only one element.
Otherwise there's no obvious performance problems. Try to benchmark and profile your program and see where the slow parts are. Usually there's one or two places that you need to optimize and get great performance. If it's still too slow, ask another question where you post your results so that we can better understand the problem.
I will join Sorin in saying don't use std::list if avoidable.
So you use key as direct index, where does it say it is none-negative? where does it say its less than 100000000?
void key_value_sequences::insert(int key, int value) {
//checks if key is valid and if the count vector needs to be resized
if(key>=0 && keyref.size() < static_cast<unsigned int>(key+1)) {
keyref.resize(key+1); // could be large
std::vector<int> val; // don't need this temporary.
seq.push_back(val); // seq is useless?
seq.back().push_back(key);
seq.back().push_back(value);
keyref[key] = seq.back(); // we now have 100000000-1 empty indexes
}
//the index is already valid
else if(key >=0) keyref[key].push_back(value);
}
Can it be done faster? depending on your key range yes it can. You will need to implement a flat_map or hash_map.
C++11 concept code for a flat_map version.
// effectively a binary search
auto key_value_sequences::find_it(int key) { // type should be iterator
return std::lower_bound(keyref.begin(), keyref.end(), [key](const auto& check){
return check[0] < key; // key is 0-element
});
}
void key_value_sequences::insert(int key, int value) {
auto found = find_it(key);
// at the end or not found
if (found == keyref.end() || found->front() != key) {
found = keyref.emplace(found, key); // add entry
}
found->emplace_back(value); // update entry, whether new or old.
}
const int* key_value_sequences::data(int key) const {
//checks if key index or ref vector is invalid
auto found = find_it(key);
if (found == keyref.end())
return nullptr;
// ->at(1) accesses the count (skipping the key) with a pointer
return found->at(1);
}
(hope I got that right ...)
I have such code:
std::unordered_map<int64_t /*id_ord*/, LimitOrder> futOrders;
auto i = futOrders.find(orderId);
if (i == futOrders.end()) {
LimitOrder& newOrder = futOrders[orderId];
// work
} else {
LimitOrder& futOrder = i->second;
// another work
}
Here I execute "find" twice:
first time: auto i = futOrders.find(orderId);
second time: LimitOrder& newOrder = futOrders[orderId];
Can i rewrite it somehow to avoid "double find"?
You can perform an emplace, and check the return value to know whether the item was inserted or not:
std::unordered_map<int64_t /*id_ord*/, LimitOrder> futOrders;
auto i = futOrders.emplace(
std::piecewise_construct, std::tie(orderId), std::make_tuple());
if (i.second) {
LimitOrder& newOrder = i.first->second;
// work
} else {
LimitOrder& futOrder = i.first->second;
// another work
}
How about using size() to realize if an element was inserted, like this:
auto old_size = futOrders.size();
LimitOrder& order = futOrders[orderId];
if (old_size < futOrders.size()) {
LimitOrder& newOrder = order;
// work
} else {
LimitOrder& futOrder = order;
// another work
}
Assuming there is a way to "determine if an order is empty", you could do:
LimitOrder& anOrder = futOrders[orderId];
if (anOrder.empty())
{
// New order, do stuff that only new orders need.
}
else
{
// Old order, update it.
}
The empty method could of course be something like if (anOrder.name == "") or if (anOrder.orderId == 0), etc.
You can use this overload of insert instead:
std::pair<iterator,bool> insert( const value_type& value );
Example:
std::unordered_map<int, std::string> m { {0, "A"}, {1, "B"}, {2, "C"} };
int orderId = 1;
// attempt to insert with key you have and default constructed value type
auto p = m.insert( std::make_pair(orderId, std::string()) );
if (p.second) {
// the element was inserted
} else {
// the element was not inserted
std::cout << p.first->second; // will print "B"
}
In both cases, p.first is the iterator to the element you search for (or just got inserted).
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);
}