ofstream或ostream类型如何将所有类型转换为字符串? [英] how does ofstream or ostream type cast all types to string?

问题描述

过去到ostream对象的任何系统定义的用户类型都将转换为字符串或char *?

any system defined user type past to ostream object is converted to a string or char* ?

like cout<<< 4<""Hello World";

like cout<<4<<"Hello World";

可以很好地工作，这是如何实现的?是<<每种类型的运算符重载?有没有一种方法可以仅通过一个通用的重载函数来实现?我的意思是我可以只使用一个带有一个参数的重载运算符方法(例如void *)，然后在该方法内部决定如何将整数类型转换为char *

works perfectly fine, how is this achieved? is the << operator overloaded for each and every type? is there a way to achieve it through just one generic overloaded function? what i mean is can i have just one overloaded operator method with one parameter(like void*) and then decide inside that method how to typecast integer to char*

如果我重载了运算符<<使用模板即

Things worked partially if i overload operator << using Template i.e

class UIStream
{
private:
 ofstream stream;
public:
 UIStream();
 ~UIStream();
 template <typename T>
 UIStream& operator << (const T);
};

这可行

 UIStream my_stream;
 my_stream<<"bcd"<10;

但是当我这样做时它会给编译器带来错误

however it gives compiler error when i do this

my_stream <<endl;

错误C2678:二进制'<<' :找不到使用"UIStream"类型的左操作数(或没有可接受的转换)的运算符

error C2678: binary '<<' : no operator found which takes a left-hand operand of type 'UIStream' (or there is no acceptable conversion)

std :: endl也不也是一种对象吗?

Is not std::endl a type of object too?

推荐答案

重新阅读您的问题后(​​由于对此此处)，而是转发到内部的流中.

After re-reading your question (as a result of a comment in this answer) I have realized that what you want is not only conversions into string (my assumptions in the other answer here), but rather forwarding into the internal ofstream.

现在，您想要实现的目标并不简单，并且在大多数情况下可能过于矫kill过正.在我拥有的[make_string][3]的实现中(转发到内部的ostringstream)，我不允许传递操纵器.如果用户想添加新行(我们在Linux下开发)，则只需传递一个'\ n'字符即可.

Now, what you want to achieve is not simple, and may be overkill in most cases. In the implementation of [make_string][3] that I have (that forwards to an internal ostringstream), I don't allow for manipulators to be passed. If the user wants to add a new line (we develop under linux) they just pass a '\n' character.

您的问题是转发操纵器(std::hex，std::endl ...).您的运算符<<被定义为采用类型T的常量实例，但是操纵器是函数指针，并且编译器无法将其与您的方法进行匹配.

Your problem is forwarding manipulators (std::hex, std::endl...). Your operator<< is defined as taking a constant instance of a type T, but manipulators are function pointers and the compiler is not able to match it against your methods.

操纵器是在std::basic_ostream模板上运行的功能. basic_ostream模板和ostream类定义为:

Manipulators are functions that operate on the std::basic_ostream template. The basic_ostream template and ostream class are defined as:

template <typename TChar, typename TTraits = char_traits<TChar> >
class basic_ostream;

typedef basic_ostream<char> ostream;
// or
// typedef basic_ostream<wchar_t> if using wide characters

然后可以传递给std :: ostream的可能的操纵器是:

Then the possible manipulators that can be passed to a std::ostream are:

typedef std::ostream& (*manip1)( std::ostream& );

typedef std::basic_ios< std::ostream::char_type, std::ostream::traits_type > ios_type;
typedef ios_type& (*manip2)( ios_type& );

typedef std::ios_base& (*manip3)( std::ios_base& );

如果要接受操纵器，则必须在类中提供该重载:

If you want to accept manipulators you must provide that overload in your class:

class mystream
{
//...
public:
   template <typename T> 
   mystream& operator<<( T datum ) {
      stream << datum;
      return *this
   }
   // overload for manipulators
   mystream& operator<<( manip1 fp ) {
      stream << fp;
      return *this;
   }
   mystream& operator<<( manip2 fp ) {
      stream << fp;
      return *this;
   }
   mystream& operator<<( manip3 fp ) {
      stream << fp;
      return *this;
   }
};

特别是，endl的签名(可能是您唯一需要的签名)是:

In particular, the signature for endl (which may be the only one you require) is:

template <typename Char, typename Traits>
std::basic_ostream<Char,Traits>& 
   std::endl( std::basic_ostream<Char,Traits>& stream );

，因此它属于manip1函数类型.其他诸如std::hex属于不同类别(在这种情况下为manip3)

so it falls under the manip1 type of functions. Others, like std::hex fall under different categories (manip3 in this particular case)

这篇关于ofstream或ostream类型如何将所有类型转换为字符串?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！