C++ 在c+；中使用向量时，如何调用构造函数和析构函数+；
我正在尝试下面的一个简单代码C++ 在c+；中使用向量时，如何调用构造函数和析构函数+；,c++,vector,constructor,destructor,C++,Vector,Constructor,Destructor,我正在尝试下面的一个简单代码 #include <iostream> #include <string> #include <vector> using namespace std; class employee { private: int emp_id; public: void getEmpid(){cout<<emp_id<<endl;} void setEmpid
#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;
}
#包括
#包括
#包括
使用名称空间std；
班级员工
{
私人：
国际环境署；
公众：
void getEmpid（）{cout对代码稍作修改将使输出易于理解。具体来说，更改析构函数：
~employee(){cout<<"destructor " << emp_id <<endl;} // shows emp_id when destructing
我得到的结果如下，我将对其进行拆分并解释：
关于向量的一些定义：
- 容量：当前分配的存储器中可容纳的元素数
- 大小：向量中的元素数
代码：员工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
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]);
}
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
a.push_back(20);
a.push_back(30);
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
调用三个构造函数，2employee（int）
和1个default。数组的前两个元素调用employee（int）
。第三个元素是默认构造的
代码：
creating array of employee[3]...
construct 2 employee id 1
construct 2 employee id 2
construct 1 employee id 10
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]);
}
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
a.push_back(20);
a.push_back(30);
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
代码：
creating array of employee[3]...
construct 2 employee id 1
construct 2 employee id 2
construct 1 employee id 10
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]);
}
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
a.push_back(20);
a.push_back(30);
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
输出：
creating array of employee[3]...
construct 2 employee id 1
construct 2 employee id 2
construct 1 employee id 10
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]);
}
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
a.push_back(20);
a.push_back(30);
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
全部完成，现在将逐个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：尽可能地使用。
此外，仔细观察，看看它的作用。阅读文档了解更多信息
- 关于病媒增长的链接：
欢迎使用StackOverflow！如果您在每个函数中打印出员工ID，可能会有助于解释发生了什么。不太清楚构造函数和析构函数是如何调用的，以及调用顺序--您应该减少main
程序，以便得到一个更易于管理的示例。您的事情太多了在main
中打开。例如，与vector
无关的第一行，employee s[8]={1,2,3,4,5}；
生成输出。你为什么不自己尝试一下呢？使用调试器，在涉及类对象的每个语句上设置断点，查看输出屏幕，观察变量。提出你的理论部分，如果你错了，我们会很乐意纠正你。确实尝试过，但不确定分配/解除分配n当容量已满时，感谢Waqar的回答非常感谢@Waqar，不确定容量已满时的分配/解除分配。