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emplace_back() 和 push_back 的区别
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"; } }
emplace_back() 和 push_back 的区别
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