What is the default allocator GCC uses for STL?

Question

According to this link, gcc provides lots of interesting memory allocators to be used with STL containers, but which is used by default if I don't specify one when creating a std::list?

Answer 1

As it says on that page you link to,

The current default choice for allocator is __gnu_cxx::new_allocator.

I.e, the default allocator is basically just operator new.

Answer 2

As per wiki :"The default allocator uses operator new to allocate memory.[13] This is often implemented as a thin layer around the C heap allocation functions,[14] which are usually optimized for infrequent allocation of large memory blocks"

from "ISO/IEC (2003). ISO/IEC 14882:2003(E): Programming Languages - C++" (wiki reference)

Default Allocator:

namespace std {   
  template <class T> class allocator;  
  // specialize for void: template <> class allocator<void>   
  {  
   public:  
   typedef void*    pointer;   
   typedef const void* const_pointer;
   // reference-to-void members are impossible. typedef void value_type;  
   template <class U> struct rebind  {  typedef allocator<U> other;  };  
};


template <class T> class allocator  
{  
public:  
  typedef size_t size_type;  
  typedef ptrdiff_t difference_type;  
  typedef T* pointer;  
  typedef const T* const_pointer;  
  typedef T& reference;  
  typedef const T& const_reference;  
  typedef T template value_type;  
  template <class U> struct rebind { typedef allocator<U> other;   
}; 

  allocator() throw();  
  allocator(const allocator&) throw();  
  template <class U> allocator(const allocator<U>&) throw();  
  ̃allocator() throw();
   pointer address(reference x) const;      
  const_pointer address(const_reference x) const;`    

  pointer allocate(  
     size_type, allocator<void>::const_pointer hint = 0);  
     void deallocate(pointer p, size_type n);  
     size_type max_size() const throw();  
     void construct(pointer p, const T& val);  
     void destroy(pointer p);  
     };  
}