如何将无参数构造函数的对象放置到std :: map中？ [英] How to emplace object with no-argument constructor into std::map?

问题描述

我想将一个对象放置到 std :: map 中，该对象的构造函数不接受任何参数。但是， std :: map :: emplace 除了键之外，似乎还需要至少一个附加参数。那么如何将零参数转发给构造函数？

I want to emplace an object into a std::map whose constructor does not take any arguments. However, std::map::emplace seems to require at least one additional argument besides the key. So how can I forward zero arguments to the constructor?

推荐答案

std :: map<的元素类型; K，V> 实际上是 std :: pair< K，V> ，因此当您嵌入到地图中时，参数将是转发给 std :: pair 的构造函数。这就是为什么您不能仅传递键的原因： std :: pair< K，V> 不能由单个参数构造（除非它是另一对相同的参数）类型。）您可以传递零个参数，但是键将被值初始化，这可能不是您想要的。

The element type of std::map<K, V> is actually std::pair<K, V>, so when you are emplacing into a map, the arguments will be forwarded to the constructor of std::pair. That's why you can't pass just the key: std::pair<K, V> can't be constructed from a single argument (unless it's another pair of the same type.) You can pass zero arguments, but then the key will be value-initialized, which is probably not what you want.

在大多数情况下，移动值将是价格便宜（并且密钥很小且可复制），您实际上应该只执行以下操作：

In most cases, moving values will be cheap (and keys will be small and copyable) and you should really just do something like this:

M.emplace(k, V{});

其中 V 是映射类型。它将被值初始化并移入容器。 （此举甚至可能会被忽略；我不确定。）

where V is the mapped type. It will be value-initialized and moved into the container. (The move might even be elided; I'm not sure.)

如果您无法移动，并且确实需要 V 要就地构造，必须使用分段构造构造器...

If you can't move, and you really need the V to be constructed in-place, you have to use the piecewise construction constructor...

M.emplace(std::piecewise_construct, std::make_tuple(k), std::make_tuple());

这会导致 std :: pair 的构造第一个元素使用 k ，第二个元素使用零参数（值初始化）。

This causes std::pair to construct the first element using k and the second element using zero arguments (value-initialization).

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