std::map emplace without copying value
Solution 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"));
Solution 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).
Comments
-
Drew Noakes almost 2 years
The C++11
std::map<K,V>
type has anemplace
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 ofkey
andval
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