Sorry for naive questions, is it OK to insert an empty set to another set using range function or it is an undefied behavior ?
Test run in https://ideone.com/RNGIFT seems fine, checking the reference saying
If the container is empty, the returned iterator will be equal to end().
#include <iostream>
#include <set>
using namespace std;
int main() {
std::set<string> to_be_inserted;
std::set<string> res;
cout << "check everything is fine" << endl;
res.insert(to_be_inserted.begin(), to_be_inserted.end());
cout << "how about now ?" << endl;
return 0;
}
Yes, most things in C++ relating to iterators will work this way in edge cases such as empty containers so that algorithms relying on the begin and end member functions on containers do not require special code in such circumstances.
Since begin will return the end iterator in the case of the set being empty as you showed, it will effectively make a range of [end, end), which has a length of 0 (as can be checked by functions like std::distance) thus preforming no insertion operations (while also being defined behavior).
This can be seen in practice in a standard library implementation, such as libc++ here where that specific overload of insert walks down the range with a for loop which has an exit condition of the two iterators (first and last) being equal, inserting elements as it goes. In the case of passing an empty range like that to it where the first and last are equal, it'll not even enter the loop.
Related
I am trying to use std::remove_if to remove spaces from a simple string, but I am getting weird results. Could someone help me figure out what's going on?
The Code is:
#include <iostream>
#include <algorithm>
#include <string>
int main(int argc, char * argv[])
{
std::string test = "a b";
std::remove_if(test.begin(), test.end(), isspace);
std::cout << "test : " << test << std::endl;
return 0;
}
I expect this to simply print out:
test : ab
but instead I get
test : abb
Trying with another string, I get:
Input: "a bcde uv xy"
Output: "abcdeuvxy xy"
It seems like it duplicating the last "word", but sometimes adds a space. How can I just get it to remove all spaces without doing weird stuff?
std::remove_if performs removing by shifting elements; the removed elements won't be erased from the container in fact. STL algorithms don't have such privilege; only containers can remove their elements.
(emphasis mine)
Removing is done by shifting (by means of move assignment) the
elements in the range in such a way that the elements that are not to
be removed appear in the beginning of the range. Relative order of the
elements that remain is preserved and the physical size of the
container is unchanged. Iterators pointing to an element between the
new logical end and the physical end of the range are still
dereferenceable, but the elements themselves have unspecified values
(as per MoveAssignable post-condition). A call to remove is typically
followed by a call to a container's erase method, which erases the
unspecified values and reduces the physical size of the container to
match its new logical size.
You can erase the removed elements afterward (which is known as erase–remove idiom).
test.erase(std::remove_if(test.begin(), test.end(), isspace), test.end());
I have a loop like this (where mySet is a std::set):
for(auto iter=mySet.begin(); iter!=mySet.end(); ++iter){
if (someCondition){mySet.insert(newElement);}
if (someotherCondition){mySet.insert(anothernewElement);}
}
I am experiencing some strange behavior, and I am asking myself if this could be due to the inserted element being inserted "before" the current iterator position in the loop. Namely, I have an Iteration where both conditions are true, but still the distance
distance(iter, mySet.end())
is only 1, not 2 as I would expect. Is my guess about set behavior right? And more importantly, can I still do what I want to do?
what I'm trying to do is to build "chains" on a hexagonal board beween fields of the same color. I have a set containing all fields of my color, and the conditions check the color of neighboring fields, and if they are of the same color, copy this field to mySet, so the chain.
I am trying to use std::set for this because it allows no fields to be in the chain more than once. Reading the comments so far I fear I need to swich to std::vector, where append() will surely add the element at the end, but then I will run into new problems due to having to think of a way to forbid doubling of elements. I therefore am hoping for advice how to solve this the best way.
Depending on the new element's value, it may be inserted before or after current iterator value. Below is an example of inserting before and after an iterator.
#include <iostream>
#include <set>
int main()
{
std::set<int> s;
s.insert(3);
auto it = s.begin();
std::cout << std::distance(it, s.end()) << std::endl; // prints 1
s.insert(2); // 2 will be inserted before it
std::cout << std::distance(it, s.end()) << std::endl; // prints 1
s.insert(5); // 5 will be inserted after it
std::cout << std::distance(it, s.end()) << std::endl; // prints 2
}
Regarding your question in the comments: In my particular case, modifying it while iterating is basically exactly what I want, but of course I need to add averything after the current position; no you can not manually arrange the order of the elements. A new value's order is determined by comparing the new one and existing elements. Below is the quote from cppreference.
std::set is an associative container that contains a sorted set of unique objects of type Key. Sorting is done using the key comparison function Compare. Search, removal, and insertion operations have logarithmic complexity. Sets are usually implemented as red-black trees.
Thus, the implementation of the set will decide where exactly it will be placed.
If you really need to add values after current position, you need to use a different container. For example, simply a vector would be suitable:
it = myvector.insert ( it+1 , 200 ); // +1 to add after it
If you have a small number of items, doing a brute-force check to see if they're inside a vector can actually be faster than checking if they're in a set. This is because vectors tend to have better cache locality than lists.
We can write a function to do this pretty easily:
template<class T>
void insert_unique(std::vector<T>& vect, T const& elem) {
if(std::find(vect.begin(), vect.end(), elem) != vect.end()) {
vect.push_back(elem);
}
}
I am trying to have multiple iterators to a bit more complex range (using range-v3 library) -- manually implementing a cartesian product, using filter, for_each and yield. However, when I tried to hold multiple iterators to such range, they share a common value. For example:
#include <vector>
#include <iostream>
#include <range/v3/view/for_each.hpp>
#include <range/v3/view/filter.hpp>
int main() {
std::vector<int> data1{1,5,2,7,6};
std::vector<int> data2{1,5,2,7,6};
auto range =
data1
| ranges::v3::view::filter([](int v) { return v%2; })
| ranges::v3::view::for_each([&data2](int v) {
return data2 | ranges::v3::view::for_each([v](int v2) {
return ranges::v3::yield(std::make_pair(v,v2));
});
});
auto it1 = range.begin();
for (auto it2 = range.begin(); it2 != range.end(); ++it2) {
std::cout << "[" << it1->first << "," << it1->second << "] [" << it2->first << "," << it2->second << "]\n";
}
return 0;
}
I expected the iterator it1 to keep pointing at the beginning of the range, while the iterator it2 goes through the whole sequence. To my surprise, it1 is incremented as well! I get the following output:
[1,1] [1,1]
[1,5] [1,5]
[1,2] [1,2]
[1,7] [1,7]
[1,6] [1,6]
[5,1] [5,1]
[5,5] [5,5]
[5,2] [5,2]
[5,7] [5,7]
[5,6] [5,6]
[7,1] [7,1]
[7,5] [7,5]
[7,2] [7,2]
[7,7] [7,7]
[7,6] [7,6]
Why is that?
How can I avoid this?
How can I keep multiple, independent iterators pointing in various locations of the range?
Should I implement a cartesian product in a different way? (that's my previous question)
While it is not reflected in the MCVE above, consider a use case where someone tries to implement something similar to std::max_element - trying to return an iterator to the highest-valued pair in the cross product. While looking for the highest value you need to store an iterator to the current best candidate. It cannot alter while you search, and it would be cumbersome to manage the iterators if you need a copy of the range (as suggested in one of the answers).
Materialising the whole cross product is not an option either, as it requires a lot of memory. After all, the whole point of using ranges with filters and other on-the-fly transformations is to avoid such materialisation.
It seems that the resulting view stores state such that it turns out to be single pass. You can work around that by simply making as many copies of the view as you need:
int main() {
std::vector<int> data1{1,5,2,7,6};
std::vector<int> data2{1,5,2,7,6};
auto range =
data1
| ranges::v3::view::filter([](int v) { return v%2; })
| ranges::v3::view::for_each([&data2](int v) {
return data2 | ranges::v3::view::for_each([v](int v2) {
return ranges::v3::yield(std::make_pair(v,v2));
});
});
auto range1= range; // Copy the view adaptor
auto it1 = range1.begin();
for (auto it2 = range.begin(); it2 != range.end(); ++it2) {
std::cout << "[" << it1->first << "," << it1->second << "] [" << it2->first << "," << it2->second << "]\n";
}
std::cout << '\n';
for (; it1 != range1.end(); ++it1) { // Consume the copied view
std::cout << "[" << it1->first << "," << it1->second << "]\n";
}
return 0;
}
Another option would be materializing the view into a container as mentioned in the comments.
Keeping in mind the aforementioned limitation of single-pass views, it is not really hard to implement a max_element
function that returns an iterator, with the important drawback of having to compute the sequence one time and a half.
Here's a possible implementation:
template <typename InputRange,typename BinaryPred = std::greater<>>
auto my_max_element(InputRange &range1,BinaryPred &&pred = {}) -> decltype(range1.begin()) {
auto range2 = range1;
auto it1 = range1.begin();
std::ptrdiff_t pos = 0L;
for (auto it2 = range2.begin(); it2 != range2.end(); ++it2) {
if (pred(*it2,*it1)) {
ranges::advance(it1,pos); // Computing again the sequence as the iterator advances!
pos = 0L;
}
++pos;
}
return it1;
}
What is goin on here?
The entire problem here originates in the fact that std::max_element requires its arguments to be LecacyForwardIterators while the ranges created by ranges::v3::yield apparently (obviously?) only provide LecacyInputIterators. Unfortunately, the range-v3 docs do not explicitly mention the iterator categories one can expect (at least I haven't found it being mentioned). This would indeed be a huge enhancement as all standard library algorithms do explicitly state what iterator categories they require.
In the particular case of std::max_element you are not the first one to stumble over this counterintuitive requirement of ForwardIterator rather than just InputIterator, see Why does std::max_element require a ForwardIterator? for example. In summary, it does make sense, though, because std::max_element does not (despite the name suggesting it) return the max element, but an iterator to the max element. Hence, it is in particular the multipass guarantee that is missing on InputIterator in order to make std::max_element work with it.
For this reason, many other standard library functions do not work with std::max_element either, e.g. std::istreambuf_iterator which really is a pity: you just cannot get the max element from a file with the existing standard library! You either have to load the entire file into memory first, or you have to use your own max algorithm.
The standard library is simply missing an algorithm that really returns the max element rather than an iterator pointing to the max element. Such an algorithm could work with InputIterators as well. Of course, this can very easily be implemented manually, but still it would be handy to have this given by the standard library. I can only speculate why it doesn't exist. Maybe one reason is, that it would require the value_type to be copy constructable because InputIterator is not required to return references to the elements and it might be in turn counterintuitive for a max algorithm to make a copy...
So, now regarding your actual questions:
Why is this? (i.e. why does your range only return InputIterators?)
Obviously, yield creates the values on the fly. This is by design, it's the very reason why one would want to use yield: to not have to create (and thus store) the range upfront. Hence, I do not see how yield could be implemented in a way that it fulfills the multipass guarantee, especially the second bullet is giving me headaches:
If a and b compare equal (a == b is contextually convertible to true) then either they are both non-dereferenceable or *a and *b are references bound to the same object
Technically, I could imagine that one could implement yield in a way that all iterators created from one range share a common internal storage that is filled on the fly during the first traversal. Then it would be possible for different iterators to give you the same references to underlying objects. But then std::max_element would silently consume O(n²) memory (all elements of your cartesian product). So, in my opinion it's definitely better to not do this and instead make the users materialize the range themselves, so that they are aware of it happening.
How can I avoid this?
Well, as already said by metalfox, you can copy your view which would result in different ranges and thus independent iterators. Still, that wouldn't make std::max_element work. So, given the nature of yield the answer to this question, unfortunately, is: you simply cannot avoid this with yield or any other technique that creates values on the fly.
How can I keep multiple, independent iterators pointing in various locations of the range?
This is related to the previous question. Basically, this question answers itself: If you want to point independent iterators in various locations, these locations have to exist somewhere in memory. So, you need to materialize at least those elements that did once have an iterator pointing to them, which in case of std::max_element means that you have to materialize all of them.
Should I implement a cartesian product in a different way?
I can imagine many different implementations. But none of them will be able to provide both of these properties all together:
return ForwardIterators
require less than O(n²) memory
Technically, it could be possible to implement an iterator that is specialized for the usage with std::max_element, meaning that it keeps only the current max element in memory so that it can be referenced... But this would be somewhat ridiculous, wouldn't it? We cannot expect a general purpose library like range-v3 to come up with such highly specialized iterator categories.
Summary
You are saying
After all, I don't think my use case is such a rare outlier and ranges
are planned to be added to the C++20 standard - so there should be
some reasonable way to achieve this without traps...
I definitely agree that "this is not a rare outlier"! However, that doesn't necessarily imply that "there should be some reasonable way to achieve this without traps". Consider e.g. NP-hard problems. It is not a rare outlier to be facing one. Still, it is impossible (unless P=NP) to solve them in polynomial time. And in your case it is simply not possible to use std::max_element without ForwardIterators. And it is not possible to implement a ForwardIterator (as defined by the standard library) on a cartesian product without consuming O(n²) memory.
For the particular case of std::max_element I would suggest to just implement your own version that returns the max element rather than an iterator pointing to it.
However, if I understand your question correctly your concern is more general and std::max_element is just an example. So, I have to disappoint you. Even with the existing standard library some trivial things are impossible due to incompatible iterator categories (again, std::istreambuf_iterator is an existing example). So, if range-v3 happens to be added, there will just be some more of such examples.
So, finally, my recommendation is to just go with your own algorithms, if possible, and swallow the pill of materializing a view otherwise.
An iterator is a pointer to an element in the vector, in this case, it1 points to the beginning of the vector. And hence, if you are trying to point the iterator to the same location of the vector, they will be the same. However, you can have multiple iterators pointing to different locations of the vector. Hope this answers your question.
I have a vector declared as
std::vector<int> MyVector;
MyVector.push_back(5);
MyVector.push_back(6);
MyVector.push_back(7);
How do should I use it in a for loop?
By iterating it with an iterator?
for (std::vector<int>::iterator it=MyVector.begin(); it!=MyVector.end(); ++it)
{
std::cout << "Vector element (*it): " << *it << std::endl;
}
Or by its access iterator?
for (std::vector<int>::size_type i=0; i<MyVector.size(); i++)
{
std::cout << "Vector element (i) : " << MyVector.at(i) << std::endl;
}
In examples I found on internet both of them are used. Is one of them superior to the other under all conditions? If not, when should I prefer one of them over the other?
The first format is more generic format for iterating over standard library containers so it is more common and intuitive. If you need to change your container then this iterating code remains unimpacted.It will work for every standard library container type, thus it gives you more generic code.
In second format, std::vector::at() checks for the bounds each time it gets called on every iteration, so it may be a little detrimental to performance. This overhead is not present in the first format as there is no bounds checking involved.Note that same is the case with using operator[].
Note the performance lag though is not as much as you will notice it unless you are operating on a huge data.
Using std::vector's [] operator is probably faster because using std::vector::at() inside a for loop checks the vector's size twice (in the for loop and in std::vector::at()'s bounds checking).
The first method can be used in other containers and thus can help you much when you change your container type.
If you use C++11, use range-based loops.
First if you have C++11, use a range-based for:
for (auto i : MyVector)
{
std::cout << i;
}
Or BOOST_FOREACH in C++03:
BOOST_FOREACH(int& i, MyVector)
{
std::cout << i;
}
Or std::copy:
std::copy(MyVector.begin(),
MyVector.end(),
std::ostream_iterator<int>(std::cout, "\n"));
As for, the question at hand, at() checks that the index is within bounds and throws an exception if it isn't. So, do not use it unless you need that extra checking. The first way you have it is standard and works well. Some people are pedantic and even it write it like so:
for (std::vector<int>::iterator it=MyVector.begin(), end = MyVector.end(); it!= end; ++it)
{
std::cout << "Vector element (*it): " << *it << std::endl;
}
In the above I cached the end iterator instead of calling end() each loop. Whether this actually makes a performance difference or not, I don't know.
There is no "one is superior to the other" (except that you almost never
want to use at()—at() is only appropriate if there is
something you can really do to recover from the error). The use of
iterator vs. index is largely one of style, and the message you're
passing. The more idiomatic C++ way of doing things would be the
iterator, but people coming from other backgrounds (for example,
mathematicians) will find indexing more idiomatic.
There are where there is a real distinction:
The iterator idiom will work with other types of containers. This
might be relevant if there is a real possibility that you use other
containers.
The indexing idiom can use a single index for several different
containers. If you're iterating through several vector with the same
size, using the indexing idiom makes it clearer that you're accessing
the same element in each of the vector. (Again, this seems to occur
most often in mathematical applications.)
Finally, any time you're really doing random access, or calculating
the element in any way, using indexes is probably more intuitive. (In
such cases, you probably want to do the calculations in int, only
converting to size_t at the last moment.)
This is a question that goes to how BOOST_FOREACH checks it's loop termination
cout << "Testing BOOST_FOREACH" << endl;
vector<int> numbers; numbers.reserve(8);
numbers.push_back(1); numbers.push_back(2); numbers.push_back(3);
cout << "capacity = " << numbers.capacity() << endl;
BOOST_FOREACH(int elem, numbers)
{
cout << elem << endl;
if (elem == 2) numbers.push_back(4);
}
cout << "capacity = " << numbers.capacity() << endl;
gives the output
Testing BOOST_FOREACH
capacity = 8
1
2
3
capacity = 8
But what about the number 4 which was inserted half way through the loop? If I change the type to a list the newly inserted number will be iterated over. The vector push_back operation will invalidate any pointers IF a reallocation is required, however that is not happening in this example. So the question I guess is why does the end() iterator appear to only be evaluated once (before the loop) when using vector but has a more dynamic evaluation when using a list?
Under the covers, BOOST_FOREACH uses
iterators to traverse the element
sequence. Before the loop is executed,
the end iterator is cached in a local
variable. This is called hoisting, and
it is an important optimization. It
assumes, however, that the end
iterator of the sequence is stable. It
usually is, but if we modify the
sequence by adding or removing
elements while we are iterating over
it, we may end up hoisting ourselves
on our own petard.
http://www.boost.org/doc/libs/1_40_0/doc/html/foreach/pitfalls.html
If you don't want the end() iterator to change use resize on the vector rather than reserve.
http://www.cplusplus.com/reference/stl/vector/resize/
Note that then you wouldn't want to push_back but use the operator[] instead. But be careful of going out of bounds.
The question was raised in the comments as to why the Microsoft debug runtime raises an assertion during iteration over the vector but not over the list. The reason is that insert is defined differently for list and vector (note that push_back is just an insert at the end of the sequence).
Per the C++ standard (ISO/IEC 14882:2003 23.2.4.3, vector modifiers):
[on insertion], if no reallocation happens, all the iterators and references before the insertion point remain valid.
(23.2.2.3, list modifiers):
[insert] does not affect the validity of iterators and references.
So, if you use push_back (and are sure that it's not going to cause a reallocation), it's okay with either container to continue using your iterator to iterate over the rest of the sequence.
In the case of the vector, however, it's undefined behavior to use the end iterator that you obtained before the push_back.
This is a roundabout answer to the question; it's a direct answer to the discussion in the question's comments.
boost's foreach will terminate when it's iterator == numbers.end()
Be careful though, calling push_back can/will invalidate any current iterators you have.