字符串分配在C++中

Question

当我发现一些奇怪的东西时,我正在玩新的和删除操作符重载.

1. 我有:

   void* operator new(size_t size)
   {
       std::cout << "Allocating memory..." << std::endl;
   
       void* p = malloc(size);
   
       if (NULL == p)
       {
           throw std::bad_alloc();
       }
   
       return p;
   }
   

2. 当我做:

   int main()
   {
       int* x = new int(1);
       std::cout << *x << std::endl;
       delete x;
       return EXIT_SUCCESS;
   }
   

   一切都按预期工作,我得到:

   Allocating memory...
   1
   

3. 但当我这样做时:

   int main()
   {
       std::string* s = new std::string("Hello world");
       std::cout << *s << std::endl;
       delete s;
       return EXIT_SUCCESS;
   }
   

   我明白了:

   Allocating memory...
   Allocating memory...
   Hello world
   

4. 事实上,当我这样做时:

   int main()
   {
       std::string s = "Hello world";
       return EXIT_SUCCESS;
   }
   

   我还是得到了Allocating memory...!

5. 最后,我这样做:

   int main()
   {
       std::string s = "Hello world";
       std::cout << &s << std::endl;
       while (true);
   }
   

   得到类似的东西:

   $ ./test &
   [1] 8979
   Allocating memory...
   0xbfc39a68
   $ cat /proc/8979/maps | grep stack
   bfc27000-bfc3c000 ... [stack]
   

   所以现在我确定s变量是在堆栈上分配的...但那么,什么叫做new运营商？我最好的猜测是它与实际文字的内存分配有关,"Hello world"但它应该是静态内存,而且new都是关于动态内存的.

这是怎么回事？

更新

在阅读了注释并自己调试示例后,我想得出结论,一旦调用了字符串构造函数,它就会在堆上为内部实现分配内存.通过跟踪new调用可以看到:

(gdb) b 13 // that's the std::cout << "Allocating memory..." << std::endl; line
(gdb) r
... Breakpoing 1, operator new (size=16) at test.cpp:13 ...
(gdb) backtrace
#0 operator new (size=16) at main.cpp:13
#1 std::string::_Rep::_S_create(unsigned int, unsigned int, std::allocator<char> const&) () from /usr/lib/libstdc++.so.6
...

并阅读std :: string(well,basic_string.tcc)源代码:

template<typename _CharT, typename _Traits, typename _Alloc>
typename basic_string<_CharT, _Traits, _Alloc>::_Rep*
basic_string<_CharT, _Traits, _Alloc>::_Rep::
_S_create(size_type __capacity, size_type __old_capacity,
          const _Alloc& __alloc)
{

   ...

    void* __place = _Raw_bytes_alloc(__alloc).allocate(__size);
   _Rep *__p = new (__place) _Rep; // Bingo!
   __p->_M_capacity = __capacity;

   ...

}

是的.编程很酷.

Answer 1

当你写作

std::string s = "Hello world";

你正在调用构造函数string (const char* s);,这个构造函数的规范是Copies the null-terminated character sequence (C-string) pointed by s.如此构造函数分配内存来存储副本.