arithmetic operations on iterator

Let it is an iterator.

it+1 or it-x I mean arithmetic increment or decrement does NOT work for iterators of sets or multisets. You have to use ++it or - -it. You know how post in/decrement and pre in/decrement works, right? So they modifies the original it iterator.

set <int> :: iterator it = it+3; //won't work

The ability to use the + operator with iterators depends on the category of the iterator.

1. Random Access Iterators:
   • Iterators belonging to the category of random access iterators (e.g., iterators of arrays, vectors, and deque in C++) support arithmetic operations like addition and subtraction.

#include <iostream>#include <vector>int main() {
    std::vector<int> myVector = {1, 2, 3, 4, 5};
    auto it = myVector.begin() + 2; // Advance iterator by 2 positions

    std::cout << "Value: " << *it << std::endl;

    return 0;
}

1. Bidirectional and Forward Iterators:

   • For iterators of categories like bidirectional and forward iterators (e.g., iterators of lists), arithmetic operations like + are not supported.

   cppCopy code
   #include <iostream>#include <list>#include <iterator>int main() {
       std::list<int> myList = {1, 2, 3, 4, 5};
       auto it = std::next(myList.begin(), 2); // Advance iterator by 2 positions
   
       std::cout << "Value: " << *it << std::endl;
   
       return 0;
   }
   
   

Alternatively, you can use next(), prev() and advance().

https://lh3.googleusercontent.com/a/AEdFTp6wkPf-TZrak2mhDBlFKcKf3oQYcPRMLw2TU3UOpA=s96-c

how does next() and prev() work? does it itself advance or backword the iterator?.

1. std::next():

   • std::next(iterator, n) returns an iterator pointing to the element that is n positions away from the original iterator.
   • The original iterator is not modified; instead, a new iterator is created and returned.
   • If n is positive, it moves forward; if negative, it moves backward.

   Example:

   cppCopy code
   #include <iostream>#include <iterator>int main() {
       std::set<int> mySet = {1, 2, 3, 4, 5};
       auto it = mySet.begin();
   
       auto nextIterator = std::next(it, 2);
   
       std::cout << "Current: " << *it << std::endl;
       std::cout << "Next (2 positions ahead): " << *nextIterator << std::endl;
   
       return 0;
   }
   

2. std::prev():

   • std::prev(iterator, n) returns an iterator pointing to the element that is n positions before the original iterator.
   • Like std::next(), the original iterator is not modified.

   Example:

   cppCopy code
   #include <iostream>#include <iterator>int main() {
       std::set<int> mySet = {1, 2, 3, 4, 5};
       auto it = mySet.end();  // pointing to the imaginary "past-the-end" element
   
       auto prevIterator = std::prev(it, 2);
   
       std::cout << "Current: " << *it << std::endl;
       std::cout << "Previous (2 positions before): " << *prevIterator << std::endl;
   
       return 0;
   }
   
   

It's essential to be cautious and make sure that the resulting iterator is valid for the container in use. If the operation would go out of bounds, you might encounter undefined behavior.

https://lh3.googleusercontent.com/a/AEdFTp6wkPf-TZrak2mhDBlFKcKf3oQYcPRMLw2TU3UOpA=s96-c

does advance() modify the original iterator?

Yes, the std::advance() function modifies the original iterator. It moves the iterator forward or backward by a specified number of positions and updates the original iterator to point to the new position.

int main() {
    std::set<int> mySet = {1, 2, 3, 4, 5};
    auto it = mySet.begin();

    // Advance the iterator by 2 positions
    std::advance(it, 2);

    // Print the value at the new position
    std::cout << "Value after advancing: " << *it << std::endl;

    return 0;
}