Moving objects from one unordered_map to another container - c++

Question

My question is that of safety. I've searched cplusplus.com and cppreference.com and they seem to be lacking on iterator safety during std::move. Specifically: is it safe to call std::unordered_map::erase(iterator) with an iterator whose object has been moved? Sample code:
#include <unordered_map>
#include <string>
#include <vector>
#include <iostream>
#include <memory>
class A {
public:
A() : name("default ctored"), value(-1) {}
A(const std::string& name, int value) : name(name), value(value) { }
std::string name;
int value;
};
typedef std::shared_ptr<const A> ConstAPtr;
int main(int argc, char **argv) {
// containers keyed by shared_ptr are keyed by the raw pointer address
std::unordered_map<ConstAPtr, int> valued_objects;
for ( int i = 0; i < 10; ++i ) {
// creates 5 objects named "name 0", and 5 named "name 1"
std::string name("name ");
name += std::to_string(i % 2);
valued_objects[std::make_shared<A>(std::move(name), i)] = i * 5;
}
// Later somewhere else we need to transform the map to be keyed differently
// while retaining the values for each object
typedef std::pair<ConstAPtr, int> ObjValue;
std::unordered_map<std::string, std::vector<ObjValue> > named_objects;
std::cout << "moving..." << std::endl;
// No increment since we're using .erase() and don't want to skip objects.
for ( auto it = valued_objects.begin(); it != valued_objects.end(); ) {
std::cout << it->first->name << "\t" << it->first.value << "\t" << it->second << std::endl;
// Get named_vec.
std::vector<ObjValue>& v = named_objects[it->first->name];
// move object :: IS THIS SAFE??
v.push_back(std::move(*it));
// And then... is this also safe???
it = valued_objects.erase(it);
}
std::cout << "checking... " << named_objects.size() << std::endl;
for ( auto it = named_objects.begin(); it != named_objects.end(); ++it ) {
std::cout << it->first << " (" << it->second.size() << ")" << std::endl;
for ( auto pair : it->second ) {
std::cout << "\t" << pair.first->name << "\t" << pair.first->value << "\t" << pair.second << std::endl;
}
}
std::cout << "double check... " << valued_objects.size() << std::endl;
for ( auto it : valued_objects ) {
std::cout << it.first->name << " (" << it.second << ")" << std::endl;
}
return 0;
}
The reason I ask is that it strikes me that moving the pair from the unordered_map's iterator may (?) therefore *re*move the iterator's stored key value and therefore invalidate its hash; therefore any operations on it afterward could result in undefined behavior. Unless that's not so?
I do think it's worth noting that the above appears to successfully work as intended in GCC 4.8.2 so I'm looking to see if I missed documentation supporting or explicitly not supporting the behavior.

Answer 1

// move object :: IS THIS SAFE??
v.push_back(std::move(*it));
Yes, it is safe, because this doesn't actually modify the key. It cannot, because the key is const. The type of *it is std::pair<const ConstAPtr, int>. When it is moved, the first member (the const ConstAPtr) is not actually moved. It is converted to an r-value by std::move, and becomes const ConstAPtr&&. But that doesn't match the move constructor, which expects a non-const ConstAPtr&&. So the copy constructor is called instead.