std::map emplace without copying value

问题

The C++11 std::map<K,V> type has an emplace function, as do many other containers.

std::map<int,std::string> m;

std::string val {\"hello\"};

m.emplace(1, val);

This code works as advertised, emplacing the std::pair<K,V> directly, however it results in a copy of key and val taking place.

Is it possible to emplace the value type directly into the map as well? Can we do better than moving the arguments in the call to emplace?

Here\'s a more thorough example:

struct Foo
{
   Foo(double d, string s) {}
   Foo(const Foo&) = delete;
   Foo(Foo&&) = delete;
}

map<int,Foo> m;
m.emplace(1, 2.3, string(\"hello\")); // invalid

回答1:

The arguments you pass to map::emplace get forwarded to the constructor of map::value_type, which is pair<const Key, Value>. So you can use the piecewise construction constructor of std::pair to avoid intermediate copies and moves.

std::map<int, Foo> m;

m.emplace(std::piecewise_construct,
          std::forward_as_tuple(1),
          std::forward_as_tuple(2.3, "hello"));

Live demo

回答2:

In C++17 this can more easily be achieved with the try_emplace method.

map<int,Foo> m;
m.try_emplace(1, 2.3, "hello");

This addition to the standard library was covered in paper N4279 and should already be supported in Visual Studio 2015, GCC 6.1 and LLVM 3.7 (the libc++ library).

来源：https://stackoverflow.com/questions/27960325/stdmap-emplace-without-copying-value

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c++

c++11

dictionary

stl

emplace