emplace_back() 和 push_back 的区别

时间:2021-08-25 04:15:44

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)的时候,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。这样造成的问题是临时变量申请的资源就浪费。 
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。 
在这上面有进一步优化的空间就是使用emplace_back.

函数原型

template <class... Args>
void emplace_back (Args&&... args);

示例

#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(
const President& other)
: name(std::move(other.name)), country(std::move(other.country)), year(other.year)
{
std::cout
<< "I am being copy 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);
};

int main()
{
std::vector
<President> elections;
std::cout
<< "emplace_back:\n";
elections.emplace_back(
"Nelson Mandela", "South Africa", 1994); //没有类的创建

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";
}

}