Question

我收到以下错误，我不确定问题是什么

1智能感知：“std :: basic_ostream＆lt; _Elem， _Traits＆gt; :: basic_ostream（const std :: basic_ostream＆lt; _Elem，_Traits＆gt; :: _ Myt＆amp; _Right）[with _Elem = char，_Traits = std :: char_traits]“（声明在“C：\ Program Files（x86）\ Microsoft Visual Studio”的第82行 11.0 \ VC \ include \ ostream“）无法访问

Book.cpp

ostream operator<< (ostream& out, const Book & b){
    out << "Title: " << b.my_Title << endl;
    out << "Author: " << b.my_Author << endl;
    out << "Number of time checkout out: " << b.my_NumberOfTimesCheckedOut;
    return(out);
}

我遇到了return(out);

的问题

Book.h

#ifndef BOOK_H
#define BOOK_H
#include <string>
using namespace std;
namespace CS20A
{
    class Book {
    public:
        Book();
        Book( string author, string title );
        string getTitle() const;
        string getAuthor() const;
        int getNumberOfTimesCheckedOut() const;
        void increaseNumberOfTimesCheckedOut( int amount=1 );
        friend ostream operator<< ( ostream& out, const Book & b );
    private:
        string my_Author;
        string my_Title;
        int my_NumberOfTimesCheckedOut;
    };
};
#endif

我甚至不明白错误告诉我的是什么

Answer 1

我怀疑你正在使用一个古老的编译器，通过使其复制构造函数成为私有来实现禁止复制std::ostream的禁止复制的std::ostream;因此令人困惑的“难以接近的”错误。

ostream &operator<< (ostream& out, const Book & b){不可复制。您必须返回参考：

{{1}}

Answer 2

我认为你的意思是返回ostream的引用。

ostream& operator<< (ostream& out, const Book & b){
    out << "Title: " << b.my_Title << endl;
    out << "Author: " << b.my_Author << endl;
    out << "Number of time checkout out: " << b.my_NumberOfTimesCheckedOut;
    return(out);
}

甚至更好，你在新的C ++版本中得到像Java这样的to_string方法。

在C ++中无法访问std :: basic_ostream

2 个答案: