问题:

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.

回答1:

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...

回答2:

Because all your implementation of std::string stores is a pointer to the heap where all of it's data is stored.

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