调用std :: map :: emplace（）并避免不必要的构造

问题描述

我有一个

选项B向前移动 str ，并将 std :: move（value）返回到对的构造函数。

所以是的，选项A构造2对，而选项B只构造1。

I have a std::map whose keys are std::string and values are my own defined type.

Let's suppose I have the following code:

std::map<std::string, MyType> mymap;
std::string str1("test");
MyType value(pars); //I want value to be moved

mymap.emplace(std::make_pair(str1, std::move(value))); //A
mymap.emplace(str, std::move(value)); //B

Assuming std::map stores pairs, I guess A would generate a further call to std::pair constructor (make_pair), followed by another call to std::pair move constructor (in-place construction with rvalue argument).

And I think B would just generate a call to std::pair constructor.

So can we say B is preferred over A in order to avoid unnecessary constructions?

解决方案

According to http://www.cplusplus.com/reference/map/map/emplace/:

So in option A, you first construct a pair which emplace will forward to the constructor (as an rvalue) for pair which will then do a move construction.

Option B forwards str and the return of std::move(value) to the constructor for pair.

So yes, option A constructs 2 pairs while option B only constructs 1.

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