在c++中使用向量时如何调用构造函数和析构函数
How does the constructors and destructors are called when using vectors in c++
我正在尝试下面的简单代码
#include <iostream>
#include <string>
#include <vector>
using namespace std;
class employee
{
private:
int emp_id;
public:
void getEmpid(){cout<<emp_id<<endl;}
void setEmpid(){ cin>>emp_id;}
employee():emp_id(10){cout<<"construct 1 "<<"employee id "<<emp_id<< endl;}
employee(int empid):emp_id(empid){cout<<"construct 2 "<<"employee id "<<emp_id<<endl;}
employee(const employee &emp):emp_id(emp.emp_id){cout<<"copy construct 3 "<<"employee id "<<emp_id<<endl;}
employee(employee&& other) : emp_id(other.emp_id) {cout<<"move construct 4 "<<"employee id "<<emp_id<<endl;}
~employee(){cout<<"destructor"<<endl;}
};
int main()
{
vector<employee>a;
employee s[8]={1,2,3,4,5};
for(int i=0;i<sizeof(s)/sizeof(s[0]);i++)
a.push_back(s[i]);
a.push_back(20);
a.push_back(30);
a.push_back(40);
a.push_back(50);
a.push_back(60);
for(int i=0;i<a.size();i++)
a[i].getEmpid();
return 0;
}
*我得到以下输出。
不太清楚如何调用构造函数和析构函数以及调用顺序。
有人可以点灯吗？ *
输出：
构造 2 员工 id 1
构造 2 员工 ID 2
构造 2 员工 ID 3
构造 2 员工 ID 4
构造 2 员工 ID 5
构造 1 个员工 ID 10
构造 1 个员工 ID 10
构造 1 个员工 ID 10
复制构造 3 员工 ID 1
复制构造 3 员工 ID 2
复制构造 3 员工 ID 1
析构函数
复制构造 3 员工 ID 3
复制构造 3 员工 ID 1
复制构造 3 员工 ID 2
析构函数
析构函数
复制构造 3 员工 ID 4
复制构造 3 员工 ID 5
复制构造 3 员工 ID 1
复制构造 3 员工 ID 2
复制构造 3 员工 ID 3
复制构造 3 员工 ID 4
析构函数
析构函数
析构函数
析构函数
复制构造 3 员工 ID 10
复制构造 3 员工 ID 10
复制构造 3 员工 ID 10
构造 2 个员工 ID 20
移动构造 4 员工 ID 20
复制构造 3 员工 ID 1
复制构造 3 员工 ID 2
复制构造 3 员工 ID 3
复制构造 3 员工 ID 4
复制构造 3 员工 ID 5
复制构造 3 员工 ID 10
复制构造 3 员工 ID 10
复制构造 3 员工 ID 10
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
构造 2 员工 id 30
移动构造 4 员工 ID 30
析构函数
构造 2 个员工 ID 40
移动构造 4 员工 ID 40
析构函数
构造 2 个员工 ID 50
移动构造 4 员工 ID 50
析构函数
构造 2 员工 id 60
移动构造 4 员工 ID 60
析构函数
1
2
3
4
5
10
10
10
20
30
40
50
60
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
对代码稍作修改将使输出更容易理解。具体来说，更改析构函数：
~employee(){cout<<"destructor " << emp_id <<endl;} // shows emp_id when destructing
并更改您的 main 以提供更小但相似的输出。我添加了一些 couts 来分隔事物并使其在查看输出时更容易理解：
int main()
{
vector<employee>a;
std::cout << "\ncreating array of employee[3]...\n" << std::endl;
employee s[3]={1,2};
std::cout << "\nstarting loop, copy into vec a\n" << std::endl;
for(int i=0;i<sizeof(s)/sizeof(s[0]);i++) {
cout << "Vec size now: " << a.size() << " Capacity: " << a.capacity() << endl;
a.push_back(s[i]);
}
cout << "Outside, Vec size now: " << a.size() << " Capacity: " << a.capacity() << endl;
std::cout << "\ndoing push back outside loop\n" << std::endl;
a.push_back(20);
a.push_back(30);
std::cout << "\nAll done exiting...\n" << std::endl;
//removed this, since we are only talking about ctors / dtors
//reduces noise
// for(int i=0;i<a.size();i++)
// a[i].getEmpid();
return 0;
}
我得到的输出如下，我将其拆分并解释：
关于向量的一些定义：
- 容量: 当前分配的存储中可以容纳的元素数
- Size: vector中元素的个数
代码: employee s[3]={1,2};
输出：
creating array of employee[3]...
construct 2 employee id 1
construct 2 employee id 2
construct 1 employee id 10
调用了三个构造函数，2 个 employee(int) 和 1 个默认构造函数。数组的前两个元素称为 employee(int)。第三个元素是默认构造的。
代码：
for(int i=0;i<sizeof(s)/sizeof(s[0]);i++) {
cout << "Vec size now: " << a.size() << " Capacity: " << a.capacity() << endl;
a.push_back(s[i]);
}
输出：
starting loop, copy into vec a
//iteration 1
Vec size now: 0 Capacity: 0 //vec is zero initially
copy construct 3 employee id 1 //id 1 is copy constructed and pushed back
//iteration 2
Vec size now: 1 Capacity: 1 //vec is now size: 1, with 1 element
//we are doing a push back, but the size needs to grow. Vector reallocates and size becomes 2. The previous memory and the elements in that memory are deallocated/destructed.
copy construct 3 employee id 2 //push back(), copy contruct into vector
copy construct 3 employee id 1 // since we reallocated, the previous memory is gone,
//we need to add the previous element as well
destructor 1 //previous element, id: 1, now being destructed.
//iteration 3
//follows the same logic as in iteration 2. size by the end will be 3 (3 elements). Capacity will be 4. That means we can do one more push back without destroying everything and reallocating.
Vec size now: 2 Capacity: 2
copy construct 3 employee id 10
copy construct 3 employee id 1
copy construct 3 employee id 2
destructor 1
destructor 2
Outside, Vec size now: 3 Capacity: 4
代码：
a.push_back(20);
a.push_back(30);
输出：
doing push back outside loop
//remember we have capacity 4
construct 2 employee id 20 //construct id 20
move construct 4 employee id 20 //move into push_back() (because 20 is temporary, rvalue)
destructor 20 //destroy what we constructed
//capacity: 4, size: 4
construct 2 employee id 30 // construct id 30
move construct 4 employee id 30 //move into push_back()
//uh oh.. capacity was full, deallocations, destructions, reconstructions:
copy construct 3 employee id 1
copy construct 3 employee id 2
copy construct 3 employee id 10
copy construct 3 employee id 20
destructor 1
destructor 2
destructor 10
destructor 20
destructor 30 //destroy our temporary id: 30
大功告成，现在只需要一个一个地调用所有元素的析构函数。
All done exiting...
destructor 10
destructor 2
destructor 1
destructor 1
destructor 2
destructor 10
destructor 20
destructor 30
教训：尽可能使用 reserve()。
此外，查看 vector::clear(), vector::shrink_to_fit，了解它的作用。阅读文档以获取更多信息。
- SO link 向量增长：About Vectors growth
- std::vector docs at cppreference
我正在尝试下面的简单代码
#include <iostream>
#include <string>
#include <vector>
using namespace std;
class employee
{
private:
int emp_id;
public:
void getEmpid(){cout<<emp_id<<endl;}
void setEmpid(){ cin>>emp_id;}
employee():emp_id(10){cout<<"construct 1 "<<"employee id "<<emp_id<< endl;}
employee(int empid):emp_id(empid){cout<<"construct 2 "<<"employee id "<<emp_id<<endl;}
employee(const employee &emp):emp_id(emp.emp_id){cout<<"copy construct 3 "<<"employee id "<<emp_id<<endl;}
employee(employee&& other) : emp_id(other.emp_id) {cout<<"move construct 4 "<<"employee id "<<emp_id<<endl;}
~employee(){cout<<"destructor"<<endl;}
};
int main()
{
vector<employee>a;
employee s[8]={1,2,3,4,5};
for(int i=0;i<sizeof(s)/sizeof(s[0]);i++)
a.push_back(s[i]);
a.push_back(20);
a.push_back(30);
a.push_back(40);
a.push_back(50);
a.push_back(60);
for(int i=0;i<a.size();i++)
a[i].getEmpid();
return 0;
}
*我得到以下输出。 不太清楚如何调用构造函数和析构函数以及调用顺序。 有人可以点灯吗？ *
输出：
构造 2 员工 id 1
构造 2 员工 ID 2
构造 2 员工 ID 3
构造 2 员工 ID 4
构造 2 员工 ID 5
构造 1 个员工 ID 10
构造 1 个员工 ID 10
构造 1 个员工 ID 10
复制构造 3 员工 ID 1
复制构造 3 员工 ID 2
复制构造 3 员工 ID 1
析构函数
复制构造 3 员工 ID 3
复制构造 3 员工 ID 1
复制构造 3 员工 ID 2
析构函数
析构函数
复制构造 3 员工 ID 4
复制构造 3 员工 ID 5
复制构造 3 员工 ID 1
复制构造 3 员工 ID 2
复制构造 3 员工 ID 3
复制构造 3 员工 ID 4
析构函数
析构函数
析构函数
析构函数
复制构造 3 员工 ID 10
复制构造 3 员工 ID 10
复制构造 3 员工 ID 10
构造 2 个员工 ID 20
移动构造 4 员工 ID 20
复制构造 3 员工 ID 1
复制构造 3 员工 ID 2
复制构造 3 员工 ID 3
复制构造 3 员工 ID 4
复制构造 3 员工 ID 5
复制构造 3 员工 ID 10
复制构造 3 员工 ID 10
复制构造 3 员工 ID 10
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
构造 2 员工 id 30
移动构造 4 员工 ID 30
析构函数
构造 2 个员工 ID 40
移动构造 4 员工 ID 40
析构函数
构造 2 个员工 ID 50
移动构造 4 员工 ID 50
析构函数
构造 2 员工 id 60
移动构造 4 员工 ID 60
析构函数
1
2
3
4
5
10
10
10
20
30
40
50
60
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
析构函数
对代码稍作修改将使输出更容易理解。具体来说，更改析构函数：
~employee(){cout<<"destructor " << emp_id <<endl;} // shows emp_id when destructing
并更改您的 main 以提供更小但相似的输出。我添加了一些 couts 来分隔事物并使其在查看输出时更容易理解：
int main()
{
vector<employee>a;
std::cout << "\ncreating array of employee[3]...\n" << std::endl;
employee s[3]={1,2};
std::cout << "\nstarting loop, copy into vec a\n" << std::endl;
for(int i=0;i<sizeof(s)/sizeof(s[0]);i++) {
cout << "Vec size now: " << a.size() << " Capacity: " << a.capacity() << endl;
a.push_back(s[i]);
}
cout << "Outside, Vec size now: " << a.size() << " Capacity: " << a.capacity() << endl;
std::cout << "\ndoing push back outside loop\n" << std::endl;
a.push_back(20);
a.push_back(30);
std::cout << "\nAll done exiting...\n" << std::endl;
//removed this, since we are only talking about ctors / dtors
//reduces noise
// for(int i=0;i<a.size();i++)
// a[i].getEmpid();
return 0;
}
我得到的输出如下，我将其拆分并解释：
关于向量的一些定义：
- 容量: 当前分配的存储中可以容纳的元素数
- Size: vector中元素的个数
代码: employee s[3]={1,2};
输出：
creating array of employee[3]...
construct 2 employee id 1
construct 2 employee id 2
construct 1 employee id 10
调用了三个构造函数，2 个 employee(int) 和 1 个默认构造函数。数组的前两个元素称为 employee(int)。第三个元素是默认构造的。
代码：
for(int i=0;i<sizeof(s)/sizeof(s[0]);i++) {
cout << "Vec size now: " << a.size() << " Capacity: " << a.capacity() << endl;
a.push_back(s[i]);
}
输出：
starting loop, copy into vec a
//iteration 1
Vec size now: 0 Capacity: 0 //vec is zero initially
copy construct 3 employee id 1 //id 1 is copy constructed and pushed back
//iteration 2
Vec size now: 1 Capacity: 1 //vec is now size: 1, with 1 element
//we are doing a push back, but the size needs to grow. Vector reallocates and size becomes 2. The previous memory and the elements in that memory are deallocated/destructed.
copy construct 3 employee id 2 //push back(), copy contruct into vector
copy construct 3 employee id 1 // since we reallocated, the previous memory is gone,
//we need to add the previous element as well
destructor 1 //previous element, id: 1, now being destructed.
//iteration 3
//follows the same logic as in iteration 2. size by the end will be 3 (3 elements). Capacity will be 4. That means we can do one more push back without destroying everything and reallocating.
Vec size now: 2 Capacity: 2
copy construct 3 employee id 10
copy construct 3 employee id 1
copy construct 3 employee id 2
destructor 1
destructor 2
Outside, Vec size now: 3 Capacity: 4
代码：
a.push_back(20);
a.push_back(30);
输出：
doing push back outside loop
//remember we have capacity 4
construct 2 employee id 20 //construct id 20
move construct 4 employee id 20 //move into push_back() (because 20 is temporary, rvalue)
destructor 20 //destroy what we constructed
//capacity: 4, size: 4
construct 2 employee id 30 // construct id 30
move construct 4 employee id 30 //move into push_back()
//uh oh.. capacity was full, deallocations, destructions, reconstructions:
copy construct 3 employee id 1
copy construct 3 employee id 2
copy construct 3 employee id 10
copy construct 3 employee id 20
destructor 1
destructor 2
destructor 10
destructor 20
destructor 30 //destroy our temporary id: 30
大功告成，现在只需要一个一个地调用所有元素的析构函数。
All done exiting...
destructor 10
destructor 2
destructor 1
destructor 1
destructor 2
destructor 10
destructor 20
destructor 30
教训：尽可能使用 reserve()。
此外，查看 vector::clear(), vector::shrink_to_fit，了解它的作用。阅读文档以获取更多信息。
- SO link 向量增长：About Vectors growth
- std::vector docs at cppreference