initializer_list and move semantics

c++ templates c++11 move-semantics initializer-list
fredoverflow picture fredoverflow · Nov 19, 2011 · Viewed 18.5k times · Source

Am I allowed to move elements out of a std::initializer_list<T>?

#include <initializer_list>
#include <utility>

template<typename T>
void foo(std::initializer_list<T> list)
{
    for (auto it = list.begin(); it != list.end(); ++it)
    {
        bar(std::move(*it));   // kosher?
    }
}

Since std::intializer_list<T> requires special compiler attention and does not have value semantics like normal containers of the C++ standard library, I'd rather be safe than sorry and ask.

Answer

Potatoswatter picture Potatoswatter · Nov 19, 2011

No, that won't work as intended; you will still get copies. I'm pretty surprised by this, as I'd thought that initializer_list existed to keep an array of temporaries until they were move'd.

begin and end for initializer_list return const T *, so the result of move in your code is T const && — an immutable rvalue reference. Such an expression can't meaningfully be moved from. It will bind to an function parameter of type T const & because rvalues do bind to const lvalue references, and you will still see copy semantics.

Probably the reason for this is so the compiler can elect to make the initializer_list a statically-initialized constant, but it seems it would be cleaner to make its type initializer_list or const initializer_list at the compiler's discretion, so the user doesn't know whether to expect a const or mutable result from begin and end. But that's just my gut feeling, probably there's a good reason I'm wrong.

Update: I've written an ISO proposal for initializer_list support of move-only types. It's only a first draft, and it's not implemented anywhere yet, but you can see it for more analysis of the problem.

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