拷贝构造函数问题

lala

#include<iostream>
using namespace std;
class Student
{
public:
Student();
Student(const char*);
Student(Student&);
char* read() {return _name;}
~Student();
private:
char* _name;
};
Student::Student():_name(0)
{
cout<<"Student Defult Constructing called..."<<endl;
}
Student::Student(const char* name)
{
cout<<"Student Constructing called..."<<name<<endl;
_name = new char[strlen(name) + 1];
strcpy(_name,name);
}
Student::Student(Student& p)
{
cout<<"Student Copying Constrcting called... :"<<p._name<<endl;
_name = new char[strlen(p._name) + 1];
strcpy(_name,p._name);
}
Student::~Student()
{
cout<<"Student Destructing called...: "<<_name<<endl;
delete _name;
}
int main()
{
Student& a = Student("abc");
cout<<a.read()<<endl;
Student b = Student("xyz"); //为什么调用的不是拷贝构造函数呢？
cout<<b.read()<<endl;
Student c(Student("lmn"));//同上；
cout<<c.read()<<endl;
Student d("qwe");
Student e(d);
cout<<e.read()<<endl;
return 0;
}

半个灵魂

以下是引用巧克力饼干在2006-4-5 21:56:00的发言：
#include<iostream>
using namespace std;
class Student
{
public:
Student();
Student(const char*);
Student(Student&);
char* read() {return _name;}
~Student();
private:
char* _name;
};
Student::Student():_name(0)
{
cout<<"Student Defult Constructing called..."<<endl;
}
Student::Student(const char* name)
{
cout<<"Student Constructing called..."<<name<<endl;
_name = new char[strlen(name) + 1];
strcpy(_name,name);
}
Student::Student(Student& p)
{
cout<<"Student Copying Constrcting called... :"<<p._name<<endl;
_name = new char[strlen(p._name) + 1];
strcpy(_name,p._name);
}
Student::~Student()
{
cout<<"Student Destructing called...: "<<_name<<endl;
delete _name;
}
int main()
{
Student& a = Student("abc");
cout<<a.read()<<endl;

//拷贝构造函数接受的参数是一个类的对象，而你这里传递的是字符串
Student b = Student("xyz"); //为什么调用的不是拷贝构造函数呢？
cout<<b.read()<<endl;
Student c(Student("lmn"));//同上；
cout<<c.read()<<endl;
Student d("qwe");
Student e(d);
cout<<e.read()<<endl;
return 0;
}

欧阳风

Student c(Student("lmn"));//为什么调用的不是拷贝构造函数呢？
Student c=Student("lmn");
Student c("lmn");
Student c(Student("lmn"));
这三种形式的对象定义初始化是等效的，因此根据所带的参数"lmn"是char *类型，那么正好和已经定义的Student::Student(const char* name)；这个构造函数是一致的，因此调用的是此构造函数，而不是拷贝构造函数

火箭筒︻$▅

问一下 Student& a = Student("abc");
和Student a = Student("abc");这两个定义一样吗？
我是菜鸟谢谢大家了