C++ C++;复制构造器
我试图很好地掌握复制构造函数&我已经找到了这部分代码C++ C++;复制构造器,c++,copy-constructor,copy-assignment,C++,Copy Constructor,Copy Assignment,我试图很好地掌握复制构造函数&我已经找到了这部分代码 #include<iostream> using namespace std; class A1 { int data; public: A1(int i = 10) : data(i) { cout << "I am constructing an A1 with: " << i <&l
#include<iostream>
using namespace std;
class A1 {
int data;
public:
A1(int i = 10) :
data(i) {
cout << "I am constructing an A1 with: " << i << endl;
}
A1(const A1& a1) :
data(a1.data) {
cout << "I am copy constructing an A1" << endl;
}
~A1() {
cout << "I am destroying an A1 with: " << data << endl;
}
void change() {
data = data * 10;
}
};
class A2 {
int data;
public:
A2(int i = 20) :
data(i) {
cout << "I am constructing an A2 with: " << i << endl;
}
A2(const A2& a2) :
data(a2.data) {
cout << "I am copy constructing an A2" << endl;
}
~A2() {
cout << "I am destroying an A2 with: " << data << endl;
}
void change() {
data = data * 20;
}
};
class A3 {
public:
A3() {
cout << "I am constructing an A3" << endl;
}
A3(const A3& a3) {
cout << "I am copy constructing an A3" << endl;
}
~A3() {
cout << "I am destroying an A3" << endl;
}
void change() {
cout << "Nothing to change" << endl;
}
};
class A {
A1 a1;
A2 a2;
A3 a3;
public:
A() {
cout << "I am constructing an A" << endl;
}
A(const A& a) :
a1(a.a1) {
cout << "I am copy constructing an A" << endl;
}
~A() {
cout << "I am destroying an A" << endl;
}
A& operator=(const A& a) {
cout << "I am performing a stupid assignment between As" << endl;
if (this != &a)
a1 = a.a1;
return *this;
}
void change() {
a1.change();
a2.change();
a3.change();
}
};
class BigA {
A data1;
A& data2;
public:
BigA(A& a) :
data1(a), data2(a) {
cout << "I just constructed a BigA" << endl;
}
~BigA() {
cout << "I am destroying a BigA" << endl;
}
A get(int index) {
if (index == 1)
return data1;
else
return data2;
}
};
BigA volta(BigA& biga)
//BigA& volta(BigA& biga)
{
cout << "Volta ta data?" << endl;
return biga;
}
int main() {
A first;
BigA biga(first);
volta(biga).get(2).change();
return 0;
}
编辑分配运算符查询:如果我在BigA中添加此函数
void change() {
A& rdata1 = data1;
A cdata2 = data2;
}
并从main调用它:biga.change()代码>为什么不调用默认赋值运算符,而是调用复制构造函数和构造函数,我得到
I am copy constructing an A1
I am constructing an A2 with: 20
I am constructing an A3
I am copy constructing an A
编辑\u AnsweringMyOwnQuery:我刚刚发现这是由复制构造函数初始化的,而不是由赋值运算符赋值的。让我们从它开始
A第一代码>
创建对象时,其字段(非静态成员)将被初始化
“”
正在调用您的无参数构造函数版本:
I am constructing an A
BigA volta(BigA& biga)
{
cout << "Volta ta data?" << endl;
return biga;
}
当你写作时
BigA-BigA(第一)代码>
调用了一个BigA
构造函数。它引用a
对象,因此不会复制first
(在提供值时设置引用)
然后,成员初始值设定项列出时间
BigA(A& a) :
data1(a), data2(a)
而data1
的类型为A
,复制第一个对象(此处引用为A
)
一个新的对象由它自己的复制构造函数创建。首先,它为A1
调用复制构造函数
A(const A& a) :
a1(a.a1)
I am copy constructing an A1
然后,A
的a2
和a3
字段默认初始化
I am constructing an A2 with: 20
I am constructing an A3
然后执行A1
的复制构造函数主体:
I am copy constructing an A
I just constructed a BigA
让我们回到BigA
初始化。我们谈到了迄今为止的data1
初始化,以及A&data2
的时间:
BigA(A& a) :
data1(a), data2(a)
由于它是引用,并且传递引用来初始化它,所以它只是一个赋值,没有输出
BigA
构造函数(采用A&
)主体,然后执行:
I am copy constructing an A
I just constructed a BigA
现在,我们将试图弄清楚发生了什么
volta(biga).get(2.change()代码>
正在调用此函数:
I am constructing an A
BigA volta(BigA& biga)
{
cout << "Volta ta data?" << endl;
return biga;
}
函数返回类BigA
的未命名对象,因此应该调用复制构造函数
您没有提供像BigA(const-BigA&BigA)
这样的复制构造函数,因此正在调用默认复制构造函数。
它对数据1执行的顺序成员初始化代码>然后A&data2代码>
通过复制未命名对象的data1
字段初始化第一个成员,从而调用A
的复制构造函数。上面解释了此处打印的内容(请参阅:一个新的A
对象由其自己的复制构造函数创建…)
然后,get
方法以index==2运行
A get(int index) {
if (index == 1)
return data1;
else
return data2; // <--- this line is executed
最后,change
运行
void change() {
a1.change();
a2.change();
a3.change();
}
只有a3.change()
打印一些内容:
Nothing to change
关于程序终止
销毁按相反的顺序进行,最后创建的change
'd对象首先被销毁
I am destroying an A
I am destroying an A3
I am destroying an A2 with: 400
I am destroying an A1 with: 100
我正在销毁一个BigA
打印了两次,但是我刚刚构建了一个BigA
——只有一次。后者是因为您没有采用const&BigA
的BigA
的复制构造函数(上面也指出了这一点)
回答你的问题
是的,复制构造函数在这里调用是正确的A cdata2=data2因为对象CDATA 2
以前未初始化。这是一个解释得很好的案例
如果你这样修改代码
A cdata2;
cdata2 = data2;
您将看到预期的任务:
I am constructing an A1 with: 10
I am constructing an A2 with: 20
I am constructing an A3
I am constructing an A
I am performing a stupid assignment between As
感谢你的超级分析性解释,这是我能问的更多。然而,作业操作员(或者我认为是这样)出现了一个问题(我编辑了我的问题),如果你也能解释一下,我将不胜感激;有趣的是,初始化顺序不依赖于初始化列表序列,而是依赖于初始化列表序列。类可以有多个构造函数,并且唯一的析构函数应该以明确的相反顺序销毁字段。
A cdata2;
cdata2 = data2;
I am constructing an A1 with: 10
I am constructing an A2 with: 20
I am constructing an A3
I am constructing an A
I am performing a stupid assignment between As