我一直在尝试编译这段代码。它有一个叫做 books 的类,其他类型的书籍继承自它。但是,当我编译程序时,它一直说 Book is an an accessible base of Police。然后它在主要的前两个 add_book 调用下显示红线,他们在其中添加了新的警察。 我没有看到我的代码中哪里缺少访问权限?
#include <iostream>
#include <vector>
#include <string>
using namespace std;
class Book{
public:
virtual double calc_price()const;
Book(string t, string a, int pg, bool bs)
: title(t), author(a), pages(pg), bestseller(bs){}
virtual ~Book(){};
virtual void display() const;
protected:
string title;
string author;
int pages;
bool bestseller;
};
void Book::display() const {
cout << "Title: " << title << endl;
cout << "Author: " << author << endl;
cout << "Number of pages: " << pages << endl;
cout << "Bestseller: "; if(bestseller==true){ cout << "Yep"
<< endl; } else {cout << "Nope" << endl;
cout << "Price: " << calc_price() << endl; }
}
double Book::calc_price() const {
if(bestseller==true){ return (pages*0.3 + 50); }
else { return (pages*0.3); }}
class Roman : public Book {
public:
Roman(string t, string a, int pg, bool bs, bool bio)
: Book(t,a,pg,bs), biography(bio){}
virtual ~Roman();
virtual void display()const override;
protected:
bool biography;
};
void Roman::display() const{
Book::display();
cout << "Ce roman ";
if(biography==true){ cout << "is a biography. " << endl;
} else { cout << "isn't a biography. " << endl; }
}
class Police : Roman {
public:
Police(string t, string a, int pg, bool bs, bool bio)
: Roman(t,a,pg,bs,bio){}
virtual double calc_price() const {
double price;
price = Book::calc_price() - 10;
if(price < 0) { return 1; } else { return price; }}
virtual~Police();
};
class Fantasy : public Livre {
public:
Fantasy(string t, string a, int pg, bool bs)
: Book(t,a,pg,bs){}
virtual ~Fantasy();
virtual double calc_price() const {
return (Book::calc_price() + 30); }
};
class Library{
public:
void display() const;
void add_book(Book* l);
void empty_stock();
private:
vector<Book*> books;
};
void Library::add_book(Book* b){
books.push_back(b);
}
void Library::display() const {
for(size_t i(0); i < books.size(); ++i){
books[i]->display();
} }
void Library::empty_stock(){
for(size_t i(0); i < books.size(); ++i){
delete books[i]; } books.clear();
}
int main()
{
Library l;
l.add_book(new Police("Dogs of Baskerville", "A.C.Doyle", 221, false,false));
l.add_book(new Police("Le Parrain ", "A.Cuso", 367, true, false));
l.add_book(new Roman("Book3", "I. Calvino", 283, false, false));
l.add_book(new Roman ("Geronimoe memories", "S.M. Barrett", 173, false, true));
l.add_book(new Fantasy ("European rivers", "C. Osborne", 150, false));
l.display();
l.empty_stock();
return 0;
}
最佳答案
更改 class Police : Roman
在 class Police : public Roman
.
如果public
未指定,Roman
将是 private
Police
的基类.
关于c++ - 不同类的不可访问基面向对象编程c++,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/29656286/