超載輸入流++什麼是錯在我的解決方案

Question

istream & read_person(istream & in, person * & p){ 
    char start, end, s, n, t, e = '\0'; 
    string surname, name, tele, email, pers =""; 
    if ((in >> start)&&(start =='<')) { 
    if((in >> pers >> s >> surname >> n >> name >> end) && (start == '<' &&  pers == "person" && s == 'S' && n == 'N' && end == '>')) { 
    person p_0(name,surname); 
    p = &p_0; 
} 
else if((in >> pers >> s >> surname >> n >> name >> t >> tele >> end) && (start == '<' && pers == "person" && s == 'S' && n == 'N' && t == 'T' && end == '>') 
{ 
    person_with_telephone p_t(name, surname, tele); 
    p = &p_t; 
} 
else if((in >> pers >> s >> surname >> n >> name >> e >> email >> end) && (start == '<' && pers == "person" && s == 'S' && n == 'N' && e == 'E' && end == '>')) 
{ 
    person_with_email p_e(name, surname, email); 
    p = &p_e; 
} 
else if((in >> pers >> s >> surname >> n >> name >> t >> tele >> e >> email >> end) && (start == '<' && pers == "person" && s == 'S' && t== 'T' && e == 'E' && n == 'N' && end == '>')) 
{ 
    person_with_telephone_and_email p_t_e(name, surname, tele, email); 
    p = &p_t_e; 
    } 
    else 
    { 
     in.setstate(ios::badbit); //read failed 

    } 
} 
return in; 
} 

這裏是人類，我有這是很容易理解它的名字和姓氏兩個字符串和一個打印方法

class person 
{ 
    string name; 
    string surname; 

    public: 

    person(){} 

    person(string strName, string strSurname):name(strName),surname(strSurname) { }   

    void set_name(string strName) 
    { name= strName; } 

    void set_surname(string strSurname) 
    { surname= strSurname; } 

    string get_name()const 
    { return name; } 

    string get_surname()const 
    { return surname; } 

    virtual bool has_telephone_p()const 
    { return false; } 

    virtual bool has_email_p()const 
    { return false; } 


    virtual void print(ostream& out)const 
    { 
      out << "<person S "<<get_surname()<<" N " << get_name() << ">"; 
    }  

    }; 

我我試圖超載iostream，可以閱讀爲這種格式 它讀取罰款的第一個，但沒有閱讀其餘的案件

Answer 1

請不要使用指針這一點。

person p_0(name,surname); // temporary person 
p = &p_0; // assign its address to a pointer 

p_0最終超出範圍。你留下了一個無效的指針和未定義的行爲。

更改operator>>簽名本：

istream &read_person(istream &in, person &p); // no pointer here, just reference 

和內部調整代碼：

p = p_0;     // making a copy 
p = person(name, surname); // you might as well do it in one line 
p = {name, surname};  // or like this 

不要朝着將來容易出錯的人慣用的方法傾斜了。