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Home/ Questions/Q 9078629
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Editorial Team
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Editorial Team
Asked: June 16, 20262026-06-16T19:40:20+00:00 2026-06-16T19:40:20+00:00

trying to learn more about how the standard library is actually implemented I’m inspecting

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trying to learn more about how the standard library is actually implemented I’m inspecting all containers in visual studio.. Here I see some curious structure:

In some base class of a std::list<> The following typedef is found

typedef typename _Alloc::template rebind<_Ty>::other _Alty;

Where “_Alloc” corresponds with the allocator template argument (and _Ty the contained type). I have trouble finding a good explanation of this “keyword”. Best thing I’ve found so far is that it is part of the allocator interface. Though even cppreference isn’t very good in explaining this.

What does this template rebind<> do? And why is it necessary at that location?

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  1. Editorial Team
    Editorial Team
    2026-06-16T19:40:21+00:00Added an answer on June 16, 2026 at 7:40 pm

    The _Alloc template is used to obtain objects of some type. The container may have an internal need to allocate objects of a different type. For example, when you have a std::list<T, A>, the allocator A is meant to allocate objects of type T but the std::list<T, A> actually needs to allocate objects of some node type. Calling the node type _Ty, the std::list<T, A> needs to get hold of an allocator for _Ty objects which is using the allocation mechanism provided by A. Using

    typename _A::template rebind<_Ty>::other
    

    specifies the corresponding type. Now, there are a few syntactic annoyances in this declaration:

    1. Since rebind is a member template of _A and _A is a template argument, the rebind becomes a dependent name. To indicate that a dependent name is a template, it needs to be prefixed by template. Without the template keyword the < would be considered to be the less-than operator.
    2. The name other also depends on a template argument, i.e., it is also a dependent name. To indicate that a dependent name is a type, the typename keyword is needed.
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