[英]Why is the copy constructor called here?
Main.cc
#include <iostream>
using namespace std;
#include <vector>
#include "Student.h"
int main()
{
Student matilda("100567899", "Matilda");
Student joe("100234555", "Joe");
Student stanley("100234888", "Stanley");
Student timmy("100234888", "Timmy");
vector<Student*> comp2404;
comp2404.push_back(&matilda);
comp2404.push_back(&joe);
comp2404.push_back(&stanley);
comp2404.push_back(&timmy);
vector<Student> vect2;
vect2.push_back(matilda);
vect2.push_back(timmy);
cout<<"all done"<<endl;
return 0;
}
Student.cc
#include <iostream>
using namespace std;
#include <string>
#include "Student.h"
Student::Student(string nu, string na)
: number(nu), name(na)
{
cout<<"-- Student default ctor "<<name<<endl;
}
Student::Student(const Student& stu)
{
name = stu.name;
number = stu.number;
cout<<"-- Student copy ctor "<<name<<endl;
}
Student::~Student()
{
cout<<"-- Student dtor"<<endl;
}
string Student::getName() const { return name; }
void Student::setName(string n) { name = n; }
ostream& operator<<(ostream& output, Student& stu)
{
output<<"Student: "<<stu.number<<" "<<stu.name<<endl;
return output;
}
輸出為:
-- Student default ctor Matilda
-- Student default ctor Joe
-- Student default ctor Stanley
-- Student default ctor Timmy
-- Student copy ctor Matilda
-- Student copy ctor Timmy
-- Student copy ctor Matilda
-- Student dtor
all done
-- Student dtor
-- Student dtor
-- Student dtor
-- Student dtor
-- Student dtor
-- Student dtor
為什么vect2
調用復制構造函數? 向量不能僅存儲實際對象(Matilda和Timmy)嗎? 另外,為什么在vect2
兩次調用了“ Matilda”?
即使vector中的push_back方法接收到一個引用,它也需要復制參數傳遞的值,否則,請考慮在訪問“ v”之前局部變量“ matilda”超出范圍時在這種情況下會發生什么情況:
int main()
{
std:vector<Student> v;
{
Student matilda("100567899", "Matilda");
v.push_back(matilda);
}
// Try to access the first element of 'v' here:
std::cout << v[0].getName() << std::endl;
}
這就是為什么它調用復制構造函數。
現在,每次std :: vector需要增加其內部數組的大小時,它將把元素從一個數組復制到另一個數組,然后再次調用復制構造函數。 這就是為什么Matilda被第二次復制。 (嘗試注釋行vect2.push_back(timmy);
並查看Matilda如何僅復制一次)。
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