Vector is another name for dynamic arrays. It is the name used for the dynamic array data structure in C++. If you have experience in Java you may know them with the name ArrayList. (Java also has an old collection class called Vector that is not used nowadays because of problems in how it was … Read more
They both store data in a linear contiguous array where accessing or iterating is both an O(1) operation so there’s no difference in performance. The only case when vector is slower is probably for some small lists because array is stored on stack in the current stack frame, hence their data is highly probably already … Read more
For the specific example you’ve given: dividing an (l,m,n) array by (m,) you can use np.newaxis: a = np.arange(1,61, dtype=float).reshape((3,4,5)) # Create a 3d array a.shape # (3,4,5) b = np.array([1.0, 2.0, 3.0, 4.0]) # Create a 1-d array b.shape # (4,) a / b # Gives a ValueError a / b[:, np.newaxis] # The … Read more
No. The elements of a vector are stored in a dynamically allocated block of memory; otherwise, the capacity of the vector could not increase. The vector object just holds a pointer to that block. The requirement that the elements be stored sequentially applies only to the elements themselves, and not to any dynamically allocated members … Read more
Your code has undefined behavior. In void set(){ X = 3; cout << “Before, X = ” << X << endl; nodes.push_back(Node()); cout << “After, X = ” << X << endl; } The access to X is really this->X and this is a pointer to the member of the vector. When you do nodes.push_back(Node()); … Read more
You need to be careful because erase() will invalidate existing iterators. However, it will return a new valid iterator you can use: for ( it = Entities.begin(); it != Entities.end(); ) { if( (*it)->getXPos() > 1.5f ) { delete * it; it = Entities.erase(it); } else { ++it; } }
Yes, this is perfectly safe to swap vectors of the same type. Vector under the hood is just a few pointers pointing to the data the vector uses and “end” of the sequence. When you call swap you just exchange those pointers between the vectors. You don’t need to worry that the vectors are the … Read more
I would expect: template<typename T> void pop_front(std::vector<T>& vec) { assert(!vec.empty()); vec.front() = std::move(vec.back()); vec.pop_back(); } to be the most efficient way of doing this, but it does not maintain the order of the elements in the vector. If you need to maintain the order of the remaining elements in vec, you can do: template<typename T> … Read more