The Allocator concept and std::allocator_traits do not say what allocate
will do when allocation fail -- will it returns nullptr or throw?
When I\'m wri
Draft n4659 for C++ standard says at 23.10.9 The default allocator [default.allocator] (emphasize mine):
23.10.9.1 allocator members [allocator.members]
...T* allocate(size_t n);
2 Returns: A pointer to the initial element of an array of storage of size n * sizeof(T), aligned appropriately for objects of type T.
3 Remarks: the storage is obtained by calling ::operator new (21.6.2), but it is unspecified when or how often this function is called.
4 Throws: bad_alloc if the storage cannot be obtained.
It makes it clear that the standard allocator will raise a bad_alloc
exception if it cannot allocate storage.
Above is for the standard allocator. The requirement for any allocator are described in 20.5.3.5 Allocator requirements [allocator.requirements] and table 31 — Allocator requirements contains:
a.allocate(n) [Return type:] X::pointer [Assertion/note/ pre-/post-condition]Memory is allocated for n objects of type T but objects are not constructed. allocate may throw an appropriate exception
My understanding is that allocate
can return only when memory has been allocated. So the allocator should throw an appropriate exception (not necessarily bad_alloc
even if it would be quite appropriate) if memory could not be allocated.