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C++中push_back和emplace_back区别


背景和区别

​emplace_back()​​ 是 ​​C++11​​ 之后,​​vector​​容器中添加的新方法,和 ​​push_back()​​一样,都是在容器末尾添加一个新的元素,相对于push_back函数,它减少了一次类的构造。不同的是​​emplace_back()​​ 在效率上相比较于 ​​push_back()​​ 有了一定的提升。

废话不多说,上经典代码:

#include <vector>
#include <string>
#include <cassert>
#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";
auto& ref = elections.emplace_back("Nelson Mandela", "South Africa", 1994);
assert(ref.year == 1994 && "uses a reference to the created object (C++17)");

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";
}
}

C++中push_back和emplace_back区别_push_back看运行结果:

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.

C++中push_back和emplace_back区别_C++ 11_02push_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。

​emplace_back()​​​ 函数在原理上比 ​​push_back()​​ 有了一定的改进,包括在内存优化方面和运行效率方面。内存优化主要体现在使用了就地构造(直接在容器内构造对象,不用拷贝一个复制品再使用)+强制类型转换的方法来实现,在运行效率方面,由于省去了拷贝构造过程,因此也有一定的提升。

emplace_back能完全代替push_back吗?

std::vector<std::vector<int>> v;
v.push_back({1,2,3}); // OK
v.emplace_back({1,2,3}); // error
v.emplace_back(std::vector<int>{1,2,3}); // OK
v.emplace_back<std::vector<int>>({1,2,3}); // OK


std::vector<std::regex> v;
v.push_back(nullptr); // 编译出错
v.emplace_back(nullptr); // 通过编译,但运行时抛出异常并且难以定位

C++中push_back和emplace_back区别_C++ 11_03


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