我正试图在几个子类中重载operator<<
。我有一个名为Question的超类,它有一个枚举值类型和一个字符串问题。这个类的子类是TextQuestion,ChoiceQuestion、BoolQuestion和ScaleQuestion。TextQuestion没有其他数据字段ChoiceQuestion有一个字符串向量,用于存储多项选择的可能性BoolQuestion没有其他数据字段ScaleQuestion有两个int值,low_和high_。
class Question {
public:
enum Type{TEXT, CHOICE, BOOL, SCALE};
Question():
type_(), question_() {}
Question(Type type, std::string question):
type_(type), question_(question) {}
friend std::ostream& operator<<(std::ostream& out, const Question& q);
virtual void print(std::ostream& out) const;
virtual ~Question();
private:
Type type_;
std::string question_;
};
class TextQuestion: public Question {
public:
TextQuestion():
Question() {}
TextQuestion(Type type, std::string question):
Question(type, question) {}
void print(std::ostream& out) const;
virtual ~TextQuestion();
};
class ChoiceQuestion: public Question {
public:
ChoiceQuestion():
Question(), choices_() {}
ChoiceQuestion(Type type, std::string question, std::vector<std::string> choices):
Question(type, question), choices_(choices) {}
void print(std::ostream& out) const;
virtual ~ChoiceQuestion();
private:
std::vector<std::string> choices_;
};
class BoolQuestion: public Question {
public:
BoolQuestion():
Question() {}
BoolQuestion(Type type, std::string question):
Question(type, question) {}
void print(std::ostream& out) const;
virtual ~BoolQuestion();
};
class ScaleQuestion: public Question {
public:
ScaleQuestion():
Question(), low_(), high_() {}
ScaleQuestion(Type type, std::string question, int low = 0, int high = 0):
Question(type, question), low_(low), high_(high) {}
void print(std::ostream& out) const;
virtual ~ScaleQuestion();
private:
int low_, high_;
};
现在,我正在尝试重载运算符<lt;对于所有这些子类,我尝试使用这个例子
所以我在超类中创建了一个虚拟print函数,重载了每个子类中的print函数,超类中的operator<<
调用print函数。
std::ostream& operator<<(std::ostream& out, const Question& q) {
q.print(out);
return out;
}
void Question::print(std::ostream& out) const {
std::string type;
switch(type_) {
case Question::TEXT:
type = "TEXT";
break;
case Question::CHOICE:
type = "CHOICE";
break;
case Question::BOOL:
type = "BOOL";
break;
case Question::SCALE:
type = "SCALE";
break;
}
out << type << " " << question_;
}
void TextQuestion::print(std::ostream& out) const {
Question::print(out);
}
void ChoiceQuestion::print(std::ostream& out) const {
Question::print(out);
out << std::endl;
int size(get_choices_size());
for (int i = 0; i < size; ++i) {
out << choices_[i] << std::endl;
}
}
void BoolQuestion::print(std::ostream& out) const {
Question::print(out);
}
void ScaleQuestion::print(std::ostream& out) const {
Question::print(out);
out << " " << low_ << " " << high_;
}
我和示例中完全一样,但当我输出Questions时,它总是使用基类,只输出类型和问题。编译器从不使用子类。
-
只有
virtual
函数允许根据对象的动态类型(给定基类的静态类型)进行调度。 -
只有非静态成员函数才能是
virtual
。 -
operator <<
不能是virtual
,因为它不能是非静态成员函数,因为它的第一个参数必须是流。 -
您可以从接受基类引用的
operator <<
中调用虚拟函数。从非虚拟函数调用虚拟函数,以保留未重写的接口方面,称为非虚拟习语。
啊,我在第一次阅读时没有看到任何virtual
,但现在我看到了。如果你没有得到预期的子类行为,一个可能的罪魁祸首是切片,你创建了一个基类类型的对象,并从子类对象中为其赋值。
TextQuestion q( "What is hello, world?" ); // Original object
Question & qr( q ); // Reference, not another object
Question q2( q ); // Base class object with copied subset (slice) of data.
std::cout << q << 'n'; // Observe subclass behavior.
std::cout << qr << 'n'; // Observe subclass behavior due to dynamic typing.
std::cout << q2 << 'n'; // Observe superclass behavior due to slicing.
不能重写类中的operator<<
,因为它不能是成员(不能将this
作为正确的操作数传递)。
相反,在每个子类中覆盖一个虚拟的print
函数,并根据它定义全局/nanespace范围operator<<
。