拷贝构造函数的参数拷贝构造函数
class A{升迁
A(const A& a);
};
问题：
参数是否⼀定要加const？
1. 拷贝构造函数的参数
1.1 使⽤不加const的拷贝构造函数
#include <iostream>
class A {
public:
A(){std::cout << "A construct." << std::endl;}
~A(){std::cout << "A desconstruct." << std::endl;}
A(A& a){std::cout << "A copy construct" << std::endl;}
//A(const A& a){std::cout << "A const copy construct" << std::endl;}
};
int main(int argc, char** argv) {
广字组词A a1;
A a2 = a1;
}
运⾏：
A construct.
目标计划A copy construct
A desconstruct.
A desconstruct.
1.2 使⽤加const的拷贝构造函数
#include <iostream>
class A {
public:
A(){std::cout << "A construct." << std::endl;}
初中数学解题技巧~A(){std::cout << "A desconstruct." << std::endl;}
//A(A& a){std::cout << "A copy construct" << std::endl;}
A(const A& a){std::cout << "A const copy construct" << std::endl;}
};
int main(int argc, char** argv) {
A a1;
A a2 = a1;
}
运⾏：
A construct.
A const copy construct
A desconstruct.
A desconstruct.
1.3 同时存在加const和不加const的拷贝构造函数
#include <iostream>
class A {
public:
A(){std::cout << "A construct." << std::endl;}
~A(){std::cout << "A desconstruct." << std::endl;}
A(A& a){std::cout << "A copy construct" << std::endl;}
A(const A& a){std::cout << "A const copy construct" << std::endl;} };
int main(int argc, char** argv) {
A a1;
初中演讲稿
A a2 = a1;
const A a3;
A a4 = a3;
}
运⾏：
A construct.
A copy construct
A construct.
A const copy construct
A desconstruct.
A desconstruct.
A desconstruct.
A desconstruct.
1.4 只存在不加const的拷贝构造函数，异常情况1:
#include <iostream>
class A {
public:
A(){std::cout << "A construct." << std::endl;}
~A(){std::cout << "A desconstruct." << std::endl;}
A(A& a){std::cout << "A copy construct" << std::endl;}
//A(const A& a){std::cout << "A const copy construct" << std::endl;} };
int main(int argc, char** argv) {
A a1;
A a2 = a1;
const A a3;
A a4 = a3;
}
编译错误： 因为a3是常量
copy_construct.cpp: In function ‘int main(int, char**)’:
copy_construct.cpp:18:9: error: binding ‘const A’ to reference of type ‘A&’ discards qualifiers
A a4 = a3;
^
copy_construct.cpp:7:2: note: initializing argument 1 of ‘A::A(A&)’
A(A& a){std::cout << "A copy construct" << std::endl;}
1.5 只存在不加const的拷贝构造函数，异常情况2：
#include <iostream>
class A {
public:
A(){std::cout << "A construct." << std::endl;}
~A(){std::cout << "A desconstruct." << std::endl;}
A(A& a){std::cout << "A copy construct" << std::endl;}
//A(const A& a){std::cout << "A const copy construct" << std::endl;}
};
A fun() {
今天的老公return A();绣眼鸟大叫
}
int main(int argc, char** argv) {
A a1;
A a2 = a1;
A a3 = fun();
}
编译失败: 编译器⽣成的临时返回对象是常引⽤类型, 因为随后临时对象赋值给其它变量时，不需要被赋值的变量去修改它;
copy_construct.cpp: In function ‘A fun()’:
copy_construct.cpp:14:11: error: invalid initialization of non-const reference of type ‘A&’ from an rvalue of type ‘A’
return A();
^
copy_construct.cpp:7:2: note: initializing argument 1 of ‘A::A(A&)’
A(A& a){std::cout << "A copy construct" << std::endl;}
^
copy_construct.cpp: In function ‘int main(int, char**)’:穷其一生
copy_construct.cpp:21:12: error: invalid initialization of non-const reference of type ‘A&’ from an rvalue of type ‘A’
A a3 = fun();
^
copy_construct.cpp:7:2: note: initializing argument 1 of ‘A::A(A&)’
A(A& a){std::cout << "A copy construct" << std::endl;}
1.6 总结:
**1. 拷贝构造函数的参数可以是A&, 也可以是const A&; **
2.如果同时存在两个拷贝构造函数, 参数分别为A&和const A&, 那么系统会根据实参类型选择拷贝构造函数
3. 如果只使⽤⾮const参数的拷贝构造函数，那么程序在编译拷贝构造函数时需要常量, 就会导致编译失败.
**4. 通常我们只编写⼀个带有const类型的拷贝构造函数, 就可以适应所有情况了. **
