c ++ 11 operator []是否等效于地图插入？ [英] Is c++11 operator[] equivalent to emplace on map insertion?
问题描述
对于C ++ 11，以下各项之间仍然存在性能差异吗？
For C++11, is there still a performance difference between the following?
（以 std :: map< Foo，std :: vector< Bar>> 为例）
map[key] = myVector and map.emplace(key, myVector)
我没有弄清楚的部分是operator []的确切内部。到目前为止，我的理解是（当键不存在时）：
The part I'm not figuring out is the exact internal of operator[]. My understanding so far has been (when key doesn't exist):
- 创建一个新键，并将关联的空默认矢量放置在内部地图
- 返回关联的空向量的引用
- 将myVector分配给该引用？
第3点是我无法理解的部分，首先如何为引用分配新值？
The point 3 is the part I couldn't understand, how can you assign a new value to a reference in the first place?
尽管我无法对第3点进行排序，但我认为某种程度上仅需要复制/移动。假设C ++ 11足够聪明，知道它将进行一次移动操作，那么整个 []赋值是否已经比insert（）便宜？它几乎等同于emplace（）吗？ ----默认构造并移动内容，而不是直接构造具有内容的矢量？
Though I cannot sort through point 3 I think somehow there's just a copy/move required. Assuming C++11 will be smart enough to know it's gonna be a move operation, is this whole "[]" assignment then already cheaper than insert()? Is it almost equivalent to emplace()? ---- default construction and move content over, versus construct vector with content directly in place?
推荐答案
有一个很多两者之间的差异。
There are a lot of differences between the two.
如果您使用 operator [] ，则 map 将默认构造该值。 operator [] 的返回值将是此默认构造的对象，然后将使用 operator = 为其分配值。
If you use operator[], then the map will default construct the value. The return value from operator[] will be this default constructed object, which will then use operator= to assign to it.
如果使用 emplace ，地图直接使用您提供的参数构造值。
If you use emplace, the map will directly construct the value with the parameters you provide.
因此， operator [] 方法将始终使用两阶段施工。如果默认构造函数的速度很慢，或者复制/移动的构造速度比复制/移动的分配速度快，则可能会出现问题。
So the operator[] method will always use two-stage construction. If the default constructor is slow, or if copy/move construction is faster than copy/move assignment, then it could be problematic.
但是，如果提供的键已经存在，则emplace 不会替换该值。而 operator [] 后跟 operator = 将始终替换该值，无论是否存在该值。
However, emplace will not replace the value if the provided key already exists. Whereas operator[] followed by operator= will always replace the value, whether there was one there or not.
也有其他差异。如果复制/移动抛出， emplace 保证地图 不会被更改。相比之下， operator [] 将始终插入默认的构造元素。因此，如果以后的复制/移动分配失败，则 map 已被更改。该键将以默认构造的 value_type 存在。
There are other differences too. If copying/moving throws, emplace guarantees that the map will not be changed. By contrast, operator[] will always insert a default constructed element. So if the later copy/move assignment fails, then the map has already been changed. That key will exist with a default constructed value_type.
确实，性能并不是您应该考虑的第一件事关于何时决定使用哪个。您需要首先关注它是否具有所需的行为。
Really, performance is not the first thing you should be thinking about when deciding which one to use. You need to focus first on whether it has the desired behavior.
C ++ 17将提供 insert_or_assign ，效果为 map [] = v ; ，但除了 insert / emplace 的安全性。
C++17 will provide insert_or_assign, which has the effect of map[] = v;, but with the exception safety of insert/emplace.
首先如何为引用分配新值？
how can you assign a new value to a reference in the first place?
从根本上说，与分配给任何非 const 参考：
It's fundamentally no different from assigning to any non-const reference:
int i = 5;
int &j = i;
j = 30;
i == 30; //This is true.
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