Question

This question is about the specification of several functions in the C++11 Standard Library, that take their arguments as rvalue references, but do not consume them in all cases. One example is
std::unordered_set<T>::insert(T&&).

It is pretty clear, that this method will use the move constructor of T to construct the element within the container, if it does not already exist. However, what happens if the element already exists in the container? I am pretty sure there is no reason the change the object in the case. However, I didn’t find anything in the C++11 Standard supporting my claim.

Here is an example to show why this might be interesting. The following code reads lines from std::cin and remove the first occurrence of duplicate lines.

std::unordered_set<std::string> seen;
std::string line;
while (getline(std::cin, line)) {
    bool inserted = seen.insert(std::move(line)).second;
    if (!inserted) {
        /* Is it safe to use line here, i.e. can I assume that the
         * insert operation hasn't changed the string object, because 
         * the string already exists, so there is no need to consume it. */
        std::cout << line << '\n';
    }
}

Apparently, this example works with GCC 4.7. But I am not sure, if it is correct according to the standard.

Answer 1

I found this note in the standard (17.4.6.9):

[ Note: If a program casts an lvalue to an xvalue while passing that lvalue to a library function (e.g. by calling the function with the argument move(x)), the program is effectively asking that function to treat that lvalue as a temporary. The implementation is free to optimize away aliasing checks which might be needed if the argument was an lvalue. — end note ]

While it doesn’t directly answer your question it does indicate that you’ve effectively “given” the argument to the library function as a temporary so I wouldn’t rely on its value once you’ve called insert. As far as I can tell, a library implementation would be entitled to move from the parameter even if it subsequently determines that it isn’t going to keep the value in the container.