I have tried many other similar questions but none of them helped me. My problem is as following:
I have 3 vectors of pointers to my struct: vector<state*>where state is my kind of struct. What I am trying to do is to remove states from vectorCheck if they are in either vectorOpen or vectorClosed. The point is, it sometimes works fine and sometimes not. According to CodeBlocks this seems to be a problem but I have no idea to overcome this. I debugged my program step by step and at some point, state from vectorCheck is not being removed despite of the fact it is in vectorClosed.
Iterating is held by 2 for loops:
vector<state*> vectorOpen;
vector<state*>::iterator itOpen;
vector<state*> vectorClosed;
vector<state*>::iterator itClosed;
vector<state*> vectorCheck;
vector<state*>::iterator itCheck;
for(itCheck = vectorCheck.begin(); itCheck != vectorCheck.end(); itCheck++) {
for(itOpen = vectorOpen.begin(); itOpen != vectorOpen.end(); itOpen++) {
if ((*itCheck)->player->x == (*itOpen)->player->x &&
(*itCheck)->player->y == (*itOpen)->player->y &&
(*itCheck)->box[0].x == (*itOpen)->box[0].x &&
(*itCheck)->box[0].y == (*itOpen)->box[0].y) {
cout << "erasing as in open " << (*itCheck)->player->x << " " << (*itCheck)->player->y << " " << (*itCheck)->box[0].x << " " << (*itCheck)->box[0].y << endl;
vectorCheck.erase(itCheck);
}
}
}
for(itCheck = vectorCheck.begin(); itCheck != vectorCheck.end(); itCheck++) {
for(itClosed = vectorClosed.begin(); itClosed != vectorClosed.end(); itClosed++) {
if((*itCheck)->player->x == (*itClosed)->player->x &&
(*itCheck)->player->y == (*itClosed)->player->y &&
(*itCheck)->box[0].x == (*itClosed)->box[0].x &&
(*itCheck)->box[0].y == (*itClosed)->box[0].y) {
cout << "erasing as in closed " << (*itCheck)->player->x << " " << (*itCheck)->player->y << " " << (*itCheck)->box[0].x << " " << (*itCheck)->box[0].y << endl;
vectorCheck.erase(itCheck);
}
}
}
Where vectorCheck is a maximum size of 3. To explain what I mean here is the picture
Where I am talking here about states in green rectangulars (3 1 2 4). Why isn't it being removed like the state in blue rectangular (2 2 2 4)? It should be removed as this state has appeared already in vectorClosed (code above).
What am I doing wrong? This is not the first iteration of the program, it happens in like 6th or 7th loop.
Also, this is probably causing my program to crash later on.
As mentioned in my comment, the problem is that you are continuing to use an iterator to an element you erased. std::vector::erase(i) is invalidating the itCheck iterator.
We can fix this by taking advantage of C++ algorithms like std::remove_if. It may make the code appear more complex at first glance, but you'll find this style of coding lets you reuse pieces of logic, improving the readability and maintainability of your code.
To start, let's write a functor that does the equality comparison you need.
struct states_are_equal :
public std::binary_function<state const *, state const *, bool>
{
bool operator()(state const * a, state const * b) const {
return a->player->x == b->player->x &&
a->player->y == b->player->y &&
a->box[0].x == b->box[0].x &&
a->box[0].y == b->box[0].y;
}
};
Now we need a predicate that will return true if the given item is found within another container. This part admittedly may be a bit hard to follow if you are not familiar with the algorithms library.
template <typename Iterator, typename Comparer>
struct is_in_container_func :
public std::unary_function<
typename std::iterator_traits<Iterator>::value_type const &,
bool
>
{
is_in_container_func(Iterator begin, Iterator end, Comparer cmp)
: it_begin(begin), it_end(end), comparer(cmp) { }
bool operator()(argument_type i) const {
return std::find_if(it_begin, it_end, std::bind1st(comparer, i)) != it_end;
}
private:
Iterator it_begin;
Iterator it_end;
Comparer comparer;
};
// This is just a helper to allow template type deduction; its only purpose is to
// allow us to omit the types for Iterator and Comparer when constructing an
// is_in_container_func object.
template <typename Iterator, typename Comparer>
is_in_container_func<Iterator, Comparer> is_in_container(
Iterator begin, Iterator end, Comparer cmp)
{
return is_in_container_func<Iterator, Comparer>(begin, end, cmp);
}
Now we can put all of these pieces together with std::remove_if:
std::vector<state*> vectorOpen;
std::vector<state*> vectorClosed;
std::vector<state*> vectorCheck;
// Make one pass, removing elements if they are found in vectorOpen.
std::vector<state*>::iterator new_end = std::remove_if(
vectorCheck.begin(), vectorCheck.end(),
is_in_container(vectorOpen.begin(), vectorOpen.end(), states_are_equal()));
// Make another pass, removing elements if they are found in vectorClosed.
new_end = std::remove_if(
vectorCheck.begin(), new_end,
is_in_container(vectorClosed.begin(), vectorClosed.end(), states_are_equal()));
// std::remove_if just swaps elements around so that the elements to be removed are
// all together at the end of the vector, and new_end is an iterator to the first
// one. So, finally, we just need to remove the range [new_end, end()).
vectorCheck.erase(new_end, vectorCheck.end());
The erase call invalidates the iterator passed to it. It shifts the elements in the vector one place to their left, and returns an iterator to the element after the removed one. Therefore, you should not increment the iterator if the erase was executed. Like so:
for(itCheck = vectorCheck.begin(); itCheck != vectorCheck.end();) { // no increment
bool found = false;
for(itOpen = vectorOpen.begin(); itOpen != vectorOpen.end(); itOpen++) {
if ((*itCheck)->player->x == (*itOpen)->player->x &&
(*itCheck)->player->y == (*itOpen)->player->y &&
(*itCheck)->box[0].x == (*itOpen)->box[0].x &&
(*itCheck)->box[0].y == (*itOpen)->box[0].y) {
cout << "erasing as in open " << (*itCheck)->player->x << " " << (*itCheck)->player->y << " " << (*itCheck)->box[0].x << " " << (*itCheck)->box[0].y << endl;
itCheck = vectorCheck.erase(itCheck);
found = true;
break; // found element and erased it. back to outer loop
}
}
if (!found) ++itCheck; // didn't find it, need to increment
}
Related
Error Window pops up right away, the program crashes etc.
Code:
void sort_star(vector<string>& product, vector<double>& star_rating)
{
vector<string>::iterator piter;
vector<double>::iterator cycler;
piter = product.begin();
cycler = star_rating.begin();
while (piter != product.end() && cycler != star_rating.end())
{
++piter; ++cycler;
cout << "/n|" << *piter << "|";
cout << *cycler << " Stars";
}
}
Yea so I'm pretty new and don't know too much on C++. A good explanation would be much appreciated!
In your while loop you increment your iterators BEFORE using them
...
while (piter != product.end() && cycler != star_rating.end()) {
++piter; ++cycler; <--- HERE
This means two things:
you skip the first element
your iterator reaches end() which is a pointer to the PAST-END-POSITION and not a valid element see here.
FIX Increment at end of loop like so:
while (piter != product.end() && cycler != star_rating.end()) {
cout << "/n|" << *piter << "|";
cout << *cycler << " Stars";
++piter; ++cycler;
}
You are dereferencing iterators after incrementing. This will
skip the first elements to be printed
will try to dereference the std::vector::end iterator.
Obviously
std::vector::end:
Returns an iterator to the element following the last element of
the container. This element acts as a placeholder; attempting to
access it results in undefined behavior.
You could have done it in a for loop as follows:
for ( ; piter != product.end() && cycler != star_rating.end(); ++piter, ++cycler)
{ // ^^ ^^^^^^^^^^^^^^^^^^
cout << "/n|" << *piter << "|" << *cycler << " Stars";
}
In page 91 of the book Elements of Programming, Stepanov and McJones say that the concept of Iterator requires a successor function but that is not necessarily regular because
...i = j does not imply that successor(i) = successor(j)...
(see page online)
I understand the converse successor(i) = successor(j) does not imply i=j (for example in two null terminated list) and that successor function may not be defined for some inputs. But I don't understand how could it be possible that i = j can lead to successor(i) != successor(j).
What case would they be referring to? Perhaps some iterator that does random (as in aleatory) hops? or some iterator that has a hidden state and "jumps" differently than the other iterator after pointing to the same element (and comparing equal in this sense).
They immediately jump to refinements (ForwardIterator) that require a regular successor function so it is not clear to me.
Initially I thought that an input iterator can have this property. However is still difficult to me to see if this constitutes a counterexample: (within a certain implementation of STL).
#include <iostream>
#include <sstream>
#include <iterator>
#include <numeric>
#include <cassert>
using std::cout; using std::endl;
int main(){
std::istream_iterator<int> it1(std::cin); // wait for one input
std::istream_iterator<int> it2 = it1;
assert(it1 == it2);
cout << "*it1 = " << *it1 << endl;
cout << "*it2 = " << *it2 << endl;
cout << "now sucessor" << endl;
++it1; // wait for one input
++it2; // wait for another input
assert(it1 == it2); // inputs still compare equal !
cout << "*it1 = " << *it1 << endl;
cout << "*it2 = " << *it2 << endl;
assert(it1 == it2); // also here ! and yet they point to different values...
assert(*it1 == *it2); // assert fails!
}
(compiled with GCC 6.1)
Consider the type iter defined as:
struct iter { unsigned value; };
inline bool operator==(iter const& x, iter const& y) {
return x.value == y.value;
}
inline bool operator!=(iter const& x, iter const& y) {
return !(x == y);
}
auto source(iter const& x) {
return x.value;
}
iter successor(iter const&) {
std::random_device engine{};
std::uniform_int_distribution<unsigned> dist{};
return {dist(engine)};
}
IIRC, iter satisfies the requirements for EoP's Iterator concept: it is Regular, source is a regular function, successor notably is not regular.
Given two objects i and j of type iter such that i == j, it is extremely likely that successor(i) != successor(j).
An example could be a successor function that consumes a stream of data (as they mention in the book).
When you have read the i-th element, you can theoretically invoke the successor function for it only once. If you try to invoke it twice, the results are different.
Simply imagine that successor(i) reads the next element from the stream, that is the i-th+1 element. It actually means to consume it and it won't be available anymore. If you call successor(i) another time, you will get the i-th+2 element from the stream.
Thus, if the inputs are the same (i = j), you have no guarantees that the outputs are the same (successor(i) = successor(j)).
I have the following for each C++ code:
for (auto item : myVector)
{
std::cout << item;
if (item == orderBy.IsLast()) // <--- Check if this is the last element
std::cout << "(Is last element) " << std::endl;
else if (item == orderBy.IsFirst()) // <-- Check if this is the first element
std::cout << "(Is first element)" << std::endl;
}
Of course IfLast() and IfFirst() do not exist on std::vector. Is there a native std:: way to check for first and last element ?
You shouldn't use the range-based for in this case, as this kind of for "hides" the iterator, and you'd need an additional counter to keep track of the position in vector. You can simply do
for(auto it = myVector.begin(); it != myVector.end(); ++it)
{
if(it == myVector.begin()) // first element
{
// do something
}
else if(std::next(it) == myVector.end()) // last element
{
// do something else
}
}
Note that simply comparing my.Vector.back() with your element from a range-based for is OK only if you're sure that you don't have duplicates in the vector. But if e.g. the value of the last element appears multiple times in the vector, you're going to find only its first position. So that's why there's really no good way of using a range-based for without an additional index that keeps track of where exactly in the vector you are.
EDIT See also #thelink2012's answer for how to "trick" your range-based for so you can get the position of the element implicitly.
Use the std::vector::front and std::vector::back to get a reference to the data in the first and last positions.
Reference is a keyword here because you could efficiently check the address of your iterating item and the address of the respective front/back references. In your example you take the item by value not reference so this prehaps wouldn't work, take in consideration this example that'd work with this method:
for(auto& item : myVector) // take item by reference
{
std::cout << item;
if (&item == &myVector.back())
std::cout << "(last element) " << std::endl;
else if (&item == &myVector.front())
std::cout << "(first element)" << std::endl;
}
If the object overloads the address of operator & (though it's considered a bad practice) you might want to use std::addressof instead.
This method won't work however for the std::vector<bool> specialization since it optimizes the vector to store booleans efficiently with bits, and since we cannot have references to bits, all references taken out this data structure is a proxy object not exactly tied to the address of the internal data.
Use std::vector::front() for the first element.
Use std::vector::back() for the last element.
Before you call those functions, make sure that the vector is not empty.
if (!orderBy.empty() && item == orderBy.back()) <--- Check if this is the last element
else if (!orderBy.empty() && item == orderBy.front()) <-- Check if this is the first element
For value based comparison, you may use myVector.front()/myVector[0] as the first and myVector.back()/myVector[myVector.size()-1] as the last element.
Suggestion
Capture the reference by default to avoid unwanted copies. e.g.
for(const auto& I : myVector)
You could search for it again, but that would be rather inefficient. If you need information on the position of the current item, you probably want to use an iterator or an index:
for (std::size_t i=0; i<myVector.size(); ++i)
{
auto& item = myVector[i];
std::cout << item;
if (i == (myVector.size() - 1))
std::cout << "(Is last element) " << std::endl;
else if (i == 0)
std::cout << "(Is first element)" << std::endl;
}
If you have special cases for the boundaries you should use the ol' iterator version, but separating the first and last cases away from the loop.
If the cases share the code after that if you should encapsulate it on a function.
I can't write code from my phone :c
This works for me
vector<int> vi {1,2,3,4};
cout << "vi = {";
for (const auto &e : vi) {
cout << e;
if (&e != &vi.back())
cout << ',';
}
cout << '}' << endl;
Pretty new to C++, only at it a week or so, I want to iterate through a set of nested sets and write each element in the inner set to a line in a file.
Each inner set has 3 elements and I want all three elements on the same line.
I have a set up as follows:
// Define "bigSet" and initiate as empty set "Triplets"
typedef set < set<string> > bigSet;
bigSet Triplets;
I tried something of this sort to go through it but it gives me an error...
// Iterate through and print output
set <string>::iterator it;
for(it = Triplets.begin(); it != Triplets.end(); it++){
cout << *it << endl;
}
Any help is greatly appreciated guys thank you!
I would do it this way:
// Iterate through and print output
set < set <string> >::iterator it_ex; // iterator for the "outer" structure
set <string>::iterator it_in; // iterator for the "inner" structure
for(it_ex = Triplets.begin(); it_ex != Triplets.end(); it_ex++)
{
for(it_in = it_ex->begin(); it_in != it_ex->end(); it_in++)
cout << *it_in << ", ";
cout << endl;
}
Triplets is not a set<string>; it is a set<set<string>>; each item in Triplets is itself a set, than can contain several strings.
The iterator must match the type of the container; with two levels of nested containers, you should iterate twice:
set<set<string>>::iterator it;
set<string>::iterator it2;
for(it = Triplets.begin(); it != Triplets.end(); it++) {
for (it2 = it->begin(); it2 != it->end(); ++it2) {
cout << *it2 << endl;
}
}
Triplets is type set < set<string> > and therefore requires an iterator of type set < set<string> >::iterator or bigSet::iterator. It isn't type set <string>. You could also use const_iterator.
Note that iterating Triplets gives you an iterator to another set, and not a string.
Also consider
for (const auto& i : Triplets)
{
for (const auto& j : i)
{
cout << j << endl;
}
}
You have an error because Triplets.begin() is not of type set<string>::iterator, it's set<set<string>>::iterator.
What you need to do is have two loops: one for iterating over the outer set and one for the inner.
set<set<string>>::iterator it;
for(it = Triplets.begin(); it != Triplets.end(); ++it)
{
set<string>::iterator it2;
for(it2 = it->begin(); it2 != it->end(); ++it2)
{
cout << *it2;
}
cout << endl;
}
If you use increment/decrement operators (++/--) on iterators, it might be better to use the prefix versions (++it) instead of the suffix ones (it++). This is because the suffix ones create a copy of the iterator before it is incremented (and that copy is then returned) but in cases like this, you have no need for it.
Moreover, if you're using C++11, you can use the range-based for loops and auto keyword, which simplify things a lot:
for(const auto &innerSet : Triplets)
{
for(const auto &innerSetElement : innerSet)
{
cout << innerSetElement;
}
cout << endl;
}
First: if they're triplets, are you sure that std::set is the type you
want for the inner values. Perhaps a class would be more
appropriate, in which case, you define an operator<< for the `class,
and your simple loop works perfectly. Something like:
class Triplet
{
std::string x;
std::string y;
std::string z;
public:
// Constructors to enforce that none of the entries are identical...
// Accessors, etc.
friend std::ostream& operator<<( std::ostream& dest, Triplet )
{
dest << x << ", " << y << ", " << z;
return dest;
}
};
And then to output:
for ( Triplet const& elem : Triplets ) {
std::cout << elem << std::endl;
}
Otherwise: you need to define the format you want for the output. In
particular, you'll probably want a separator between the strings in the
line, for example. Which means you probably cannot use a range based
for, at least not for the inner loop. You would need something like:
for ( std::set<std::string> const& triplet : Triplets ) {
for ( auto it = triplet.cbegin(); it != triplet.cend(); ++it ) {
if ( it != triplet.cebegin() ) {
std::cout << ", ";
}
std::cout << *it;
}
std::cout << std::endl;
}
(If the set of triplets is large, you'll definitely want to consider
replacing std::endl with '\n'. But of course, if it is really
large, you probably won't be outputting to std::cout.)
My remove_if seems to be overwriting the elements that are not filtered out with values of filtered out elements. The purpose of these code is to allow user to filter and display only teacher from a certain category. (Not deleting any element)
Here are some of the code
static string compare;
static string debug;
bool filter_Cat (Teacher &t)
{
return (t.getCat() != compare);
}
void filterCat (vector<Teacher> &t)
{
vector<Teacher>::iterator i;
vector<Teacher>::iterator newedited = remove_if(t.begin(), t.end(), filter_Cat);
for (i = t.begin(); i != newedited; ++i)
{
Teacher& te = *i;
te.getName();
cout << "\t";
te.getCategory();
cout << "\t";
te.getLocation();
}
}
void filterTutorCat(vector<Teacher> &t)
{
int choice;
cout << "No\tCategory" << endl
<< "1\tEnglish" << endl
<< "2\tMath" << endl
<< "3\tScience" << endl
<< "Choose the category you wish to filter :";
cin >> choice;
getline(cin, debug);
if(choice <= 3 && choice > 0)
{
if (choice == 1)
{
compare = "English";
filterCat(t);
}
if (choice == 2)
{
compare = "Math";
filterCat(t);
}
if (choice == 3)
{
compare = "Science";
filterCat(t);
}
}
else
{
cout << "Invalid Option" << endl;
}
}
remove_if shifts elements, for which the compare function returns false, from right to left; which in other words means, it overwrites the elements, for which compare returns true, with elements, for which compare returns false. The size of the vector doesn't change, however.
This reads,
Removes all elements satisfying specific criteria from the range [first, last). The first version removes all elements that are equal to value, the second version removes all elements for which predicate p returns true.
Removing is done by shifting the elements in the range in such a way that elements to be erased are overwritten. The elements between the old and the new ends of the range have unspecified values. Iterator to the new end of the range is returned. Relative order of the elements that remain is preserved.
So what you want to do should be expressed as:
void filterCat (vector<Teacher> &v)
{
for (vector<Teacher>::iterator it = v.begin(); it != v.end() ; ++it)
{
if (!filter_Cat(*i))
{
std::cout << i->getName() <<"\t" << i->getCategory() << std::endl;
}
}
}
It seems in your code, getName() prints the name which ideally it should not do, instead it should return name. So I would suggest you to change it to make it return name. And do the same for getCategory as well. Choose your name correctly. If it is getName(), you should get you name by returning it; if it is printName(), then it should print name.
Also, the code which you've written isn't good:
You should avoid global variables.
You should avoid if-else as much as possible. Learn better ways.
You should learn about function objects (or functor)
You should learn about const member function.
You should understand the difference between iterator and const_iterator, and their usage.
You should understand the difference between const reference, and non-const reference. And try using them appropriately.
So I would write your code as:
//this is functor, not a function
struct filter_cat
{
std::string m_cat; //use member data, avoid global variable
filter_cat(std::string const & cat) : m_cat(cat) {}
bool operator()(Teacher const & t) const //const member function
{
return (t.getCat() != m_cat); //getCat should be const member function
}
};
//pass vector by const reference
void filterCat (vector<Teacher> const & v, filter_cat filter)
{
//use const_iterator here, instead of iterator
for (vector<Teacher>::const_iterator it = v.begin(); it != v.end() ; ++it)
{
if (!filter(*i))
{
//getName and getCategory should be const member function
std::cout << i->getName() <<"\t" << i->getCategory() << std::endl;
}
}
}
void filterTutorCat(vector<Teacher> const &t)
{
int choice;
cout << "No\tCategory" << endl
<< "1\tEnglish" << endl
<< "2\tMath" << endl
<< "3\tScience" << endl
<< "Choose the category you wish to filter :";
cin >> choice;
getline(cin, debug);
//avoid if-else as much as possible, learn better ways!
std::string cats[] = {"English", "Math", "Science"};
if(choice <= 3 && choice > 0)
{
filterCat(v, filter_cat(cats[choice-1]));
}
else
{
cout << "Invalid Option" << endl;
}
}
As noted in the comments: getCat, getName and getCategory should be const member functions. In fact, if getCategory returns category, then getCat isn't even needed.
Solved my issue.
remove_if collects the values for which filter_Cat returns false at the start of the container. While it doesn't reduce the number of elements in the container it neither does make any guarantees about the values of the elements beyond the returned range. So you are loosing values when using remove_if.