为什么sizeof（std :: string）只有8个字节？ [英] Why is sizeof(std::string) only eight bytes?

问题描述

为什么 std :: string 的大小取决于 sizeof（std :: string） yield 8 ？

我认为应该超过 8 int （ sizeof（int）== 8 在我的机器上）data成员给予 std :: string :: length（）和 std :: string :: size（） std： ：string 未由C ++标准指定。它只描述类的行为。但是，我希望在类中有多个指针的信息价值。特别是：

• 指向实际字符串的指针。




• 使用的实际大小。

位置，因此占用与 char * [在大多数架构]中完全相同的空间量。

事实上，看看我的Linux机器附带的C ++头文件，当你看看（其中，根据注释，是pre-C ++ 11，但是我认为这两种方法都有代表性）：

  size_type 
 length（）const _GLIBCXX_NOEXCEPT 
 {return _M_rep（） - > _M_length; } 
  

，然后按照：

  _Rep * 
 _M_rep（）const _GLIBCXX_NOEXCEPT 
 {return&（（reinterpret_cast< _Rep *>（_M_data（）））[ -  1] } 
  

这又导致：

  _CharT * 
 _M_data（）const _GLIBCXX_NOEXCEPT 
 {return _M_dataplus._M_p; } 
  

这导致

  //数据成员（私人）：
 mutable _Alloc_hider _M_dataplus; 
  

，然后我们进入：

  struct _Alloc_hider：_Alloc 
 {
 _Alloc_hider（_CharT * __dat，const _Alloc& __a）_GLIBCXX_NOEXCEPT 
：_Alloc（__a），_M_p 
 
 _CharT * _M_p; //实际数据。 
}; 
  

字符串的实际数据为：

  struct _Rep_base 
 {
 size_type _M_length; 
 size_type _M_capacity; 
 _Atomic_word _M_refcount; 
}; 
  

所以，这是一个简单的指针 _M_p 隐藏在几层吸气剂和一点铸造...

Why is std::string's size, as determined by sizeof(std::string), yield 8?
I thought it should be more than 8 as it has to have an int (sizeof(int) == 8 on my machine) data member for giving std::string::length() and std::string::size() in O(1) and probably a char* for characters.

解决方案

The implementation of std::string is not specified by the C++ standard. It only describes the classes behaviour. However, I would expect there to be more than one pointer's worth of information in the class. In particular:

• A pointer to the actual string.
• The size available.
• The actual size used.

It MAY of course store all these in a dynamically allocated location, and thus take up exactly the same amount of space as char* [in most architectures].

In fact looking at the C++ header that comes with my Linux machine, the implementation is quite clear when you look at (which, as per comments, is "pre-C++11", but I think roughly representative either way):

  size_type
  length() const _GLIBCXX_NOEXCEPT
  { return _M_rep()->_M_length; }

and then follow that to:

  _Rep*
  _M_rep() const _GLIBCXX_NOEXCEPT
  { return &((reinterpret_cast<_Rep*> (_M_data()))[-1]); }

which in turn leads to:

  _CharT*
  _M_data() const _GLIBCXX_NOEXCEPT
  { return  _M_dataplus._M_p; }

Which leads to

  // Data Members (private):
  mutable _Alloc_hider  _M_dataplus;

and then we get to:

  struct _Alloc_hider : _Alloc
  {
    _Alloc_hider(_CharT* __dat, const _Alloc& __a) _GLIBCXX_NOEXCEPT
    : _Alloc(__a), _M_p(__dat) { }

    _CharT* _M_p; // The actual data.
  };

The actual data about the string is:

  struct _Rep_base
  {
    size_type       _M_length;
    size_type       _M_capacity;
    _Atomic_word        _M_refcount;
  };

So, it's all a simple pointer called _M_p hidden inside several layers of getters and a bit of casting...

这篇关于为什么sizeof（std :: string）只有8个字节？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！