Can assignment due to dereferencing const_iterator cause undefined behavior?

Question

This code is a simplified test for what I'm trying to do elsewhere. I have a function that takes a ref-to-ptr argument and modifies it to return a pointer from a list of pointers.

#include <iostream> #include <list> using namespace std; typedef int* intp; typedef std::list<intp> intplist; intplist myList; void func(intp &arg) // (1) { intplist::const_iterator it = myList.begin(); std::advance(it, 2); arg = *it; } int main() { myList.push_back(new int(1)); myList.push_back(new int(2)); myList.push_back(new int(3)); int* ip = NULL; // (2) func(ip); if (ip) cout << "ip = " << *ip << endl; else cout << "ip is null!" << endl; for (intplist::const_iterator it = myList.begin(); it != myList.end(); ++it) delete *it; return 0; } 

It works and prints ip = 3 , as expected, only it bothers me that it can cause undefined behavior or otherwise lead to problems, because I remove the iterator constant, assigning the result of its dereference to Argument. I tried adding const to (1) and (2) but did not create one.

Am I right to worry? If so, why am I not getting a warning from g ++ (4.9.2)?

Answer 1

The code is great. You do not remove any constant (there is no way to do this implicitly in C ++). *it gives you const intp & . You copy the pointer that this link refers to in arg . Copying from something does not separate the continent. The arg assignment is assigned to ip in your case, it does not bind anything to the intp object inside the container.

Answer 2

const_iterator simply means that you cannot assign this iterator and / or you can only call const functions on the object it points to. There is no problem copying value - in this case a pointer. You do not save constant pointers; if you were, you would need to assign a const pointer