使用聚合初始化时，是否可以在地图中进行无复制放置? [英] Is it possible to do a no-copy emplace into map while using aggregate initialization?

问题描述

有关如何插入而不复制地图值.

See this answer for how to insert into stdmap without making copies of the map value.

std :: map放置而不复制值

从该答案继续-假设我的 Foo 类型看起来像这样:

Continueing from that answer - suppose my Foo type looks something like this:

struct Foo {
  const int& intref_; 
  std::mutex mutex_;
}

然后使用像这样的聚合初始化来初始化

Then initialized using aggregate-initialization like this

Foo{7}

或

Foo{7, std::mutex()}

以某种方式可以将其放置到地图中吗?:

Would it be somehow possible to be emplaced into the map with type ?:

std::map<size_t, Foo> mymap;

我知道我可以为 Foo 写一个构造函数-但是可以用聚合初始化来代替吗?

I know I could just write a constructor for Foo - but can it be done with aggregate initialization instead ?

链接到编译器资源管理器:

https://godbolt.org/z/_Fm4k1

相关的c ++参考:

https://en.cppreference.com/w/cpp/container/map/try_emplace

https://en.cppreference.com/w/cpp/language/aggregate_initialization

推荐答案

您可以利用类型转换来间接构建您的建筑

You may exploit casts to indirect your construction

template<typename T>
struct tag { using type = T; };

template<typename F>
struct initializer
{
    F f;
    template<typename T>
    operator T() &&
    {
        return std::forward<F>(f)(tag<T>{});
    }
};

template<typename F>
initializer(F&&) -> initializer<F>;

template<typename... Args>
auto initpack(Args&&... args)
{
    return initializer{[&](auto t) {
        using Ret = typename decltype(t)::type;
        return Ret{std::forward<Args>(args)...};
    }};
}

并将其用作

struct Foo
{
  const int& intref_; 
  std::mutex mutex_;
};

void foo()
{
    int i = 42;
    std::map<int, Foo> m;
    m.emplace(std::piecewise_construct,
              std::forward_as_tuple(0),
              std::forward_as_tuple(initpack(i)));
}

请注意，您不能通过将临时目录绑定到非堆栈引用来延长其使用寿命.

Note you can't prolong a temporary's lifetime by binding it to a non-stack reference.

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