关于 C++ 中自定义 I/O 运算符的问题？

发布于 2024-11-18 11:29:45 字数 1041 浏览 1 评论 0原文

我有以下代码：

 class Student {
 public:
     Student(){}
     void display() const{}
     friend istream& operator>>(istream& is, Student& s){return is;}
     friend ostream& operator<<(ostream& os, const Student& s){return os; }
 };
int main()
{
    Student st;
    cin >> st;
    cout << st;

    getch();
    return 0;
}

我自己尝试过省略 friend 关键字，使运算符成为 Student 类的成员函数，然后编译器会生成“binary 'operator > ;>'参数太多”。我读过一些文档说发生这种情况是因为所有成员函数总是接收一个隐式参数“this”（这就是为什么所有成员函数都可以访问私有变量）。基于该解释，我尝试如下：

class Student {
 public:
     Student(){}
     void display() const{}
     istream& operator>>(istream& is){return is;}
     ostream& operator<<(ostream& os){return os; }
 };
int main()
{
    Student st;
    cin >> st;
    cout << st;

    getch();
    return 0;
}

并收到错误消息：“错误 C2679：二进制 '>>” : 找不到采用“Student”类型右侧操作数的运算符（或者没有可接受的转换）”

谁能给我一个明确的解释？

原文

I have the following piece of code:

 class Student {
 public:
     Student(){}
     void display() const{}
     friend istream& operator>>(istream& is, Student& s){return is;}
     friend ostream& operator<<(ostream& os, const Student& s){return os; }
 };
int main()
{
    Student st;
    cin >> st;
    cout << st;

    getch();
    return 0;
}

I have tried myself when omitting the friend keywords to make the operators become the member function of the Student class, then the compiler would produce "binary 'operator >>' has too many parameters". I have read some document saying that happened because all member functions always receive an implicit parameter "this" (that's why all member functions can access private variables).
Based on that explanation, I have tried as follows:

class Student {
 public:
     Student(){}
     void display() const{}
     istream& operator>>(istream& is){return is;}
     ostream& operator<<(ostream& os){return os; }
 };
int main()
{
    Student st;
    cin >> st;
    cout << st;

    getch();
    return 0;
}

And got the error message: "error C2679: binary '>>' : no operator found which takes a right-hand operand of type 'Student' (or there is no acceptable conversion)"

Can anyone give me a clear explanation, please?

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忘羡 2024-11-25 11:29:45

您不能说该函数是友元函数，然后将该函数内联包含在内。 friend关键字暗示该函数未在类中定义，但它可以访问该类的所有私有变量和受保护变量。将代码更改为：

class Student {
  public:
    Student(){}
    void display() const{}
    friend istream& operator>>(istream& is, Student& s);
    friend ostream& operator<<(ostream& os, const Student& s);
};

istream& operator >>(istream& is, Student& s) { return is; }
ostream& operator <<(ostream& os, const Student& s) { return os; }

查看 http://www.java2s.com/Code/另一个例子是 Cpp/Overload/Overloadstreamoperator.htm。

与<<和>>，左侧操作数始终是文件流，因此您不能在实际类中重载它们（从技术上讲，它必须进入文件流类）。

You can't say that the function is a friend function, and then include the function in-line. The friend keyword implies that the function is not defined in the class, but it can access all the private and protected variables of the class. Change your code to:

class Student {
  public:
    Student(){}
    void display() const{}
    friend istream& operator>>(istream& is, Student& s);
    friend ostream& operator<<(ostream& os, const Student& s);
};

istream& operator >>(istream& is, Student& s) { return is; }
ostream& operator <<(ostream& os, const Student& s) { return os; }

Look at http://www.java2s.com/Code/Cpp/Overload/Overloadstreamoperator.htm for another example.

With << and >>, the left hand operand is always a file stream, so you cannot overload them within your actual class (it'd technically have to go in the file stream class).

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