我正在尝试通过复制构造函数创建我的类的实例的深层副本,但我无法弄清楚,如何编写它。在这一刻,当我调用复制构造函数时,程序不会崩溃然而,当我想对实例做任何事情时(即打印数组,添加一些项目等),然后程序崩溃......
有人可以告诉我,如何正确地写出来?它让我疯狂但O_o
struct DbChange {
const char* date;
const char* street;
const char* city;
};
class DbPerson {
public:
DbPerson(void);
const char* id;
const char* name;
const char* surname;
DbChange * change;
int position;
int size;
};
DbPerson::DbPerson() {
position = 0;
size = 1000;
change = new DbChange[1000];
}
class Register {
public:
// default constructor
Register(void);
int size;
int position;
DbPerson** db;
//copy constructor
Register(const Register& other) : db() {
db= new DbPerson*[1000];
std::copy(other.db, other.db + (1000), db);
}
};
int main(int argc, char** argv) {
Register a;
/*
* put some items to a
*/
Register b ( a );
a . Print (); // now crashes
b . Print (); // when previous line is commented, then it crashes on this line...
return 0;
}
答案 0 :(得分:4)
由于显示的代码绝不允许我们猜测Print的作用,以及它为什么会发生冲突,我只会告诉你我是如何期待C ++中的东西(而不是C和Java之间的混淆) :
<强> http://liveworkspace.org/code/4ti5TS$0 强>
#include <vector>
#include <string>
struct DbChange {
std::string date;
std::string street;
std::string city;
};
class DbPerson {
public:
DbPerson(void);
std::string id, name, surname;
int position;
std::vector<DbChange> changes;
size_t size() const { return changes.size(); }
};
DbPerson::DbPerson() : position(), changes() { }
class Register {
public:
size_t size() const { return db.size(); }
int position; // unused?
std::vector<DbPerson> db;
Register() = default;
//copy constructor
Register(const Register& other) : db(other.db)
{
// did you forget to copy position? If so, this would have been the
// default generated copy constructor
}
void Print() const
{
// TODO
}
};
int main() {
Register a;
/*
* put some items to a
*/
Register b(a);
a.Print(); // now crashes
b.Print(); // when previous line is commented, then it crashes on this line...
return 0;
}