Question

I want to clear a element from a vector using the erase method. But the problem here is that the element is not guaranteed to occur only once in the vector. It may be present multiple times and I need to clear all of them. My code is something like this:

void erase(std::vector<int>& myNumbers_in, int number_in) {     std::vector<int>::iterator iter = myNumbers_in.begin();     std::vector<int>::iterator endIter = myNumbers_in.end();     for(; iter != endIter; ++iter)     {         if(*iter == number_in)         {             myNumbers_in.erase(iter);         }     } }  int main(int argc, char* argv[]) {     std::vector<int> myNmbers;     for(int i = 0; i < 2; ++i)     {         myNmbers.push_back(i);         myNmbers.push_back(i);     }      erase(myNmbers, 1);      return 0; } 

This code obviously crashes because I am changing the end of the vector while iterating through it. What is the best way to achieve this? I.e. is there any way to do this without iterating through the vector multiple times or creating one more copy of the vector?

Answer 1

Since C++20, there are freestanding std::erase and std::erase_if functions that work on containers and simplify things considerably:

std::erase(myNumbers, number_in); // or std::erase_if(myNumbers, [&](int x) { return x == number_in; }); 

Prior to C++20, use the erase-remove idiom:

std::vector<int>& vec = myNumbers; // use shorter name vec.erase(std::remove(vec.begin(), vec.end(), number_in), vec.end()); // or vec.erase(std::remove_if(vec.begin(), vec.end(), [&](int x) {     return x == number_in; }), vec.end()); 

What happens is that std::remove compacts the elements that differ from the value to be removed (number_in) in the beginning of the vector and returns the iterator to the first element after that range. Then erase removes these elements (whose value is unspecified).