C ++操作符重载示例 [英] C++ Operator overloading example

问题描述

好吧，我是操作员重载的新，我发现这个问题。而不是记录自己，我更喜欢问你：D

Well, I'm new to operator overloading, and I found this problem. Instead of documenting myself, I prefer to ask you :D

重点是，我知道如何做简单的操作符重载，但是我面临堆栈操作符的问题。我将尝试一个相对简单的例子：

The point is, I know how to do simple operator overloading, but I'm facing problems with stacking operators. I'll try to put a relatively simple example:

struct dxfdat
{
 int a;
 string b;

 /* here is the question */
}

/* use: */
dxfdat example;

example << "lalala" << 483 << "puff" << 1029 << endl;

lalala< 483< puff< 1029<< endl 应存储在 b 。

dxfdat&运算符<< （T a），并且类似的东西使用一个参数（例如<< 7），但我希望它工作在' cout '时尚。

dxfdat& operator<< (T a) and things like that work with one parameter (example << 7), but I would like it to work in a 'cout' fashion.

很抱歉懒惰。

EDIT：

真正的东西...好吧，它有点棘手...实际上，b不是字符串，的其他对象， example<< lalala< 483< puff< 1029<< endl 应该只创建一个对象。

The real thing... Ok, it is a little bit trickier... actually, b isn't a string, but a vector of other objects, and example << "lalala" << 483 << "puff" << 1029 << endl should just create just one object.

这是我正在尝试（翻译），虽然我没有线索如何告诉它是什么时候创建的对象（从左到右，不是吗？）：

This is what I'm trying (translated), though I have no clue on how to tell it when to create the object (as it goes from left to right, doesn't it?):

struct dxfDato
{
    dxfDato(int c = 0, string v = 0, int t = 0) { cod = c; val= v; ty = t; }

    int ty;
    int cod;
    string val;
};

struct dxfItem
{
    int cl;
    string val;
    vector<dxfDato> dats;
    vector<dxfItem> sons;

    template <class T>
    dxfItem &operator<<(const T &t)
    {
        dxfDato dd;
        std::stringstream ss;
        ss << t;
        val = ss;
        dats.push_back(dd); // this way, it creates a lot of objects
        return d;
    }

};

dxfItem headers;

headers << "lalala" << 54789 << "sdfa" << 483 << endl;
// this should create *just one object* in dats vector,
// and put everything on string val

感谢一切，

注意：我不得不提取和翻译很多东西放在这里，所以我没有检查代码中的愚蠢错误。

Note: I had to extract and translate a lot of things to put it here, so I didn't check the code for stupid errors.

（对不起，扩大的问题，请告诉我，如果我误用stackoverflow的问题系统）

(Sorry for expanding the question that much, please tell me if I'm misusing stackoverflow's question system)

推荐答案

很容易，不要惊慌：）

它非常类似于 std :: cout - std :: endl 工作。

You have recognized the problem well: it's very similar to the std::cout - std::endl work.

你可以这样做，但如果你不介意，我会重命名类型。

You could do like such, though I'll rename the types, if you don't mind.

struct EndMarker {};
extern const EndMarker end; // To be defined in a .cpp

class Data
{
public:
  Data(): m_data(1, "") {}

  // Usual operator
  template <class T>
  Data& operator<<(const T& input)
  {
    std::ostringstream aStream;
    aStream << input;
    m_data.back() += aStream.str();
  };

  // End of object
  Data& operator<<(EndMarker) { m_data.push_back(""); }

private:
  std::vector<std::string> m_data;
}; // class Data

它默认添加到当前最后一个元素，

It works by adding to the current last element by default, and pushing an empty element at the end.

让我们看一个例子：

Data data;
data << 1 << "bla" << 2 << end << 3 << "foo" << end;

// data.m_data now is
// ["1bla2", "3foo", ""]

另一个解决方案是如果 end 已经流式传输，则保存一个标志（布尔）它有，在下一次插入时创建一个新的元素（并擦除标志）。

The other solution would be to keep a flag (boolean) to store if a end has been streamed or not, and if it has, creating a new element on the next insertion (and erasing the flag).

插入一点工作，但是没有空元素...您的电话。

It a bit more work on insertion, but you don't have the empty element... your call.

这篇关于C ++操作符重载示例的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！