为什么会出现这个问题？解决这个问题的原理是什么？
#include<iostream>
using namespace std;
class Merchandise{
public:
Merchandise(){};
~Merchandise(){};
friend istream &operator >> (istream &in,Merchandise &s);
friend ostream &operator << (ostream &out,Merchandise &s);
private:
int no;
char *name;
double price;
};
istream& operator >>(istream &in,Merchandise &s)
{
cout<<"please input commoditiy's no:"<<endl;
in>>s.no;
cout<<"please input commoditiy's name:"<<endl;
in>>s.name;
cout<<"please input commoditiy's price:"<<endl;
in>>s.price;
return in;
}
ostream& operator <<(ostream &out,Merchandise &s)
{
out<<"\t"<<"commoditiy's information"<<endl;
out<<"no"<<"\t"<<"name"<<"\t"<<"price"<<endl;
out<<s.no<<"\t"<<s.name<<"\t"<<s.price<<endl;
return out;
}
int main()
{ Merchandise mer;
cin>>mer;
cout<<mer;
return 0;
}
[解决办法]
char *name;
//g改成
char name[24];
//或者在构造函数中动态分配内存空间，在析构里释放就好！
//多说一句，构造函数和析构函数后面的分号不需要，有误导作用！
#include<iostream>
using namespace std;
class Merchandise{
public:
Merchandise(){};
~Merchandise(){};
friend istream &operator >> (istream &in,Merchandise &s);
friend ostream &operator << (ostream &out,Merchandise &s);
private:
int no;
char name[50];
double price;
};
istream& operator >>(istream &in,Merchandise &s)
{
cout<<"please input commoditiy's no:"<<endl;
in>>s.no;
cout<<"please input commoditiy's name:"<<endl;
//运行到下面一行代码的时候会报异常，这是必然的，因为
//name根本就没有初始化，也没有内存地址，是0xcccccccc
//在定义char *name;的时候改为char name[50]就好了
in>>s.name;
cout<<"please input commoditiy's price:"<<endl;
in>>s.price;
return in;
}
ostream& operator <<(ostream &out,Merchandise &s)
{
out<<"\t"<<"commoditiy's information"<<endl;
out<<"no"<<"\t"<<"name"<<"\t"<<"price"<<endl;
out<<s.no<<"\t"<<s.name<<"\t"<<s.price<<endl;
return out;
}
int main()
{ Merchandise mer;
cin>>mer;
cout<<mer;
return 0;
}
