What does a declaration of template look like when using the standard allocator in a template class?

Question

I saw some examples of using allocator as a class member in a template class. Such like:

template <class T, class Alloc = std::allocator<T>> class myVector { public: ... protected: std::allocator<value_type> _alloc; ... private: ... };

But the code still works when I delete the default-value template argument like template <class T> class myVector. So do we need to add a default-value template argument when we have a allocator as class member? If the answer is yes, why?

The code you've shown does not use the Alloc argument, so deleting it doesn't change anything. — Pete Becker
– Pete Becker, Commented Feb 16, 2019 at 20:41

erenon · Accepted Answer · 2019-02-16 20:43:25Z

The code shown is probably wrong: It should use the provided Alloc type to allocate, instead of hard-coding std::allocator. (And also take advantage of the empty-base-class optimization, to avoid increasing the container size if the allocator is an empty type)

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