confusion about scope resolution operator in c++

Question

what :: does in line : return ::operator new(size, ::std::nothrow); and why the class is using template when there is no use for template type T

template<typename T>
class DefaultMemoryAllocator
{
public:
  static inline void *Allocate(size_t size)
  {
    return ::operator new(size, ::std::nothrow);
  }
  static inline void Deallocate(void *pointer, size_t size)
  {
    ::operator delete(pointer);
  }
};

It's "scope resolution operator". And the reason for the type is explained here http://stackoverflow.com/questions/4173254/what-is-the-curiously-recurring-template-pattern-crtp — StoryTeller - Unslander Monica, Jan 16 '17 at 11:23

Bathsheba · Accepted Answer · 2017-01-16T11:30:11.320

1

Using the scope resolution operator :: like this means that the global operator new and operator delete functions are called, as opposed to ones that might have been overridden for that class.

You'll probably find that this function is part of a memory policy class, and is called from the class' operator new and operator delete overridden functions.

edited Jan 16 '17 at 11:30

answered Jan 16 '17 at 11:23

Bathsheba

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why the class is using template when there is no use for template type T? – tohidprogram Jan 16 '17 at 11:27
2

@tohidprogram That might be easier to say if you shared where the class is used, and for what. – Angew is no longer proud of SO Jan 16 '17 at 11:28
It looks to me like it's a snippet from a policy type framework based on the curiously recurring template pattern (CRTP - Google that). *That* particular member function doesn't use the `T`, but other ones probably do. – Bathsheba Jan 16 '17 at 11:28

confusion about scope resolution operator in c++

1 Answers1