C++11中emplace_back和 push_back 的区别
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2022-03-21 16:28:25
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c++开发中我们会经常用到插入操作对stl的各种容器进行操作,比如vector,map,set等。在引入右值引用,转移构造函数,转移复制运算符之前,通常使用push_back()向容器中加入一个右值元素(临时对象)时,首先会调用构造函数构造这个临时对象,然后需要调用拷贝构造函数将这个临时对象放入容器中。原来的临时变量释放。这样造成的问题就是临时变量申请资源的浪费。
引入了右值引用,转移构造函数后,push_back()右值时就会调用构造函数和转移构造函数,如果可以在插入的时候直接构造,就只需要构造一次即可。这就是c++11 新加的emplace_back。
emplace_back函数原型:
template <class... Args>
void emplace_back (Args&&... args);
在容器尾部添加一个元素,这个元素原地构造,不需要触发拷贝构造和转移构造。而且调用形式更加简洁,直接根据参数初始化临时对象的成员。
最后再来一段代码,涉及到使用右值引用和std::move的
#include <vector>
#include <string>
#include <iostream>
struct President
{
std::string name;
std::string country;
int year;
President(std::string && p_name, std::string && p_country, int p_year)
: name(std::move(p_name)), country(std::move(p_country)), year(p_year)
{
std::cout << "I am being constructed.\n";
}
President(President&& other)
: name(std::move(other.name)), country(std::move(other.country)), year(other.year)
{
std::cout << "I am being moved.\n";
}
President& operator=(const President& other) = default;
};
int main()
{
std::vector<President> elections;
std::cout << "emplace_back:\n";
elections.emplace_back("Nelson Mandela", "South Africa", 1994);
std::vector<President> reElections;
std::cout << "\npush_back:\n";
reElections.push_back(President("Franklin Delano Roosevelt", "the USA", 1936));
std::cout << "\nContents:\n";
for (President const& president : elections) {
std::cout << president.name << " was elected president of "
<< president.country << " in " << president.year << ".\n";
}
for (President const& president : reElections) {
std::cout << president.name << " was re-elected president of "
<< president.country << " in " << president.year << ".\n";
}
}
//输出:
emplace_back:
I am being constructed.
push_back:
I am being constructed.
I am being moved.
Contents:
Nelson Mandela was elected president of South Africa in 1994.
Franklin Delano Roosevelt was re-elected president of the USA in 1936.
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