参见英文答案 >
vector push_back calling copy_constructor more than once?5个
我想知道为什么在下面的C代码中复制构造函数被调用25次10次迭代？
我想知道为什么在下面的C代码中复制构造函数被调用25次10次迭代？
如果它是10然后OK 10/10 = 1,或20/10 = 2,或30/10 = 3,但25/10 = 2.5？这里的.5是什么意思？
标题：
class Person
{
public:
Person(std::string name,int age);
Person(const Person &person);
const std::string &getName() const;
int getAge() const;
private:
std::string name;
int age;
};
资源：
Person::Person(string name,int age) : name(std::move(name)),age(age)
{}
Person::Person(const Person &person)
{
this->name = person.name;
this->age = person.age;
static int count = 0;
count++;
cout << ">>Copy-Person::Person(Person &person) " << count << endl;
}
const string &Person::getName() const
{
return name;
}
int Person::getAge() const
{
return age;
}
用法：
int main()
{
vector<Person> persons;
for (int i = 0; i < 10; ++i)
{
Person person(to_string(i + 1),i);
persons.push_back(person);
}
cout << "-----------------------------------------------" << endl;
for (Person &person : persons)
{
cout << "name = " << person.getName() << " age = " << person.getAge() << endl;
}
return 0;
}
输出：
>>Copy-Person::Person(Person &person) 1 >>Copy-Person::Person(Person &person) 2 >>Copy-Person::Person(Person &person) 3 >>Copy-Person::Person(Person &person) 4 >>Copy-Person::Person(Person &person) 5 >>Copy-Person::Person(Person &person) 6 >>Copy-Person::Person(Person &person) 7 >>Copy-Person::Person(Person &person) 8 >>Copy-Person::Person(Person &person) 9 >>Copy-Person::Person(Person &person) 10 >>Copy-Person::Person(Person &person) 11 >>Copy-Person::Person(Person &person) 12 >>Copy-Person::Person(Person &person) 13 >>Copy-Person::Person(Person &person) 14 >>Copy-Person::Person(Person &person) 15 >>Copy-Person::Person(Person &person) 16 >>Copy-Person::Person(Person &person) 17 >>Copy-Person::Person(Person &person) 18 >>Copy-Person::Person(Person &person) 19 >>Copy-Person::Person(Person &person) 20 >>Copy-Person::Person(Person &person) 21 >>Copy-Person::Person(Person &person) 22 >>Copy-Person::Person(Person &person) 23 >>Copy-Person::Person(Person &person) 24 >>Copy-Person::Person(Person &person) 25 ----------------------------------------------- name = 1 age = 0 name = 2 age = 1 name = 3 age = 2 name = 4 age = 3 name = 5 age = 4 name = 6 age = 5 name = 7 age = 6 name = 8 age = 7 name = 9 age = 8 name = 10 age = 9
解决方法
你没有为你的人物矢量保留任何记忆.这意味着当push_back期间的persons.size()== persons.capacity()时,向量将在堆上分配一个新的更大的缓冲区并将每个元素复制到它.这就是为什么你会看到比预期更多的副本.
如果你写…
persons.reserve(10);
…在循环之前,您将看不到任何“额外”副本.
请注意,您可以通过使用std :: vector :: emplace_back和std :: vector :: reserve来避免副本altogheter：
for (int i = 0; i < 10; ++i)
{
persons.emplace_back(to_string(i + 1),i);
}
这只会打印：
name = 1 age = 0
name = 2 age = 1
name = 3 age = 2
name = 4 age = 3
name = 5 age = 4
name = 6 age = 5
name = 7 age = 6
name = 8 age = 7
name = 9 age = 8
name = 10 age = 9
