Ostream方法不起作用

ostream method does not work

本文关键字:不起作用 方法 Ostream      更新时间:2023-10-16

我有两个方法,第一个定义字符串:

我不明白为什么它没有做到,以及我如何从main调用这个方法。

Actor::operator std::string( ) const {
    std::stringstream ss;
    ss << this->_id;
    std::string str1 = ss.str();
    std::stringstream s;
    s << this->_salary;
    std::string str2 = s.str();
    std::string str3 = "Actor first name = " + this->_firstname + ", last name = " + this->_lastname+", id = " + str1 + ", monthly salary = " + str2;
    if (this->_hasoscar==true)
        str3+=" was NOMINATED Oscar AWARD..";
    return str3;
}

下一个需要打印出来

const Actor& Actor::print(std::ostream& os) {
    os<< std::string();
    return *this;
}

不清楚为什么要这样做,因为正常的工作方式是重载ostream操作符:

class Actor  {
public:
  friend std::ostream& operator<< (std::ostream& os, const Actor& a) {
    os << "Actor first name = " + a._firstname + 
          ", last name = " + a._lastname+", id = " + 
          a._id + ", monthly salary = " + a._salary;
    if (this->_hasoscar) {
       os << " was NOMINATED Oscar AWARD.."; 
    } 
    return os;
  }
};

通过这种方式,您可以从main中轻松地创建和打印actor:

Actor a;
cout << a; 

第一个方法

Actor::operator std::string() const();

允许从Actor到字符串的隐式转换,例如:

Person a;
std::string s = a;

os<< std::string();

这不会调用到string的转换,它创建一个临时字符串变量并将其写入流,因此它相当于:

os << "";

要调用转换操作符,需要告诉编译器要将对象转换为字符串(而不仅仅是构造一个空字符串)。这样做的显式方法是:

os << static_cast<std::string>(*this);

还有其他方法,例如:

os << std::string(*this);
os << (std::string)*this;
std::string s = *this;
os << s;

或者如果你喜欢不可读的代码,你可以显式调用转换操作符:

os << this->operator std::string();

(最后一种方法不是个好主意。)

转换操作符的定义非常愚蠢,并且比需要的慢得多。使用两个单独的字符串流和字符串连接是非常浪费的,你可以用一个字符串流来完成这一切:

Actor::operator std::string( ) const {
    std::stringstream ss;
    ss << "Actor first name = " << this->_firstname
       << ", last name = " << this->_lastname
       << ", id = " << this->_id
       << ", monthly salary = " << this->_salary;
    if (this->_hasoscar==true)
        ss << " was NOMINATED Oscar AWARD..";
    return ss.str();
}

然而,正如Velthune所说,除非出于其他原因需要转换为字符串,否则输出类的通常方法是重载该类的operator<<