std::allocator_traits<Alloc>::allocate_at_least
[[nodiscard]] static constexpr std::allocation_result< pointer, size_type >
allocate_at_least( Alloc& a, size_type n );
allocate_at_least calls a.allocate_at_least(n) and returns its result if the call is well-formed, otherwise, it is equivalent to return {a.allocate(n), n};.
allocator_at_least tries to allocate a storage for at least n value_type objects, and provides a fallback mechanism that allocates a storage for exact n objects.
Parameters
a - an allocator used for allocating storage
n - the lower bound of number of objects to allocate storage for
Return value
a.allocate_at_least(n) if it is well-formed.
Otherwise, std::allocation_result<pointer, size_type>{a.allocate(n), n}.
Exceptions
Throws what and when the selected allocation function throws.
Notes
The allocate_at_least member function of Allocator types are mainly provided for contiguous containers, e.g. std::vector and std::basic_string, in order to reduce reallocation by making their capacity match the actually allocated size when possible. Because allocate_at_least provides a fallback mechanism, it can be directly used where appropriate.
Given an allocator object a of type Alloc, let result denote the value returned from std::allocator_traits<Alloc>::allocate_at_least(a, n),
the storage should be deallocated by a.deallocate(result.ptr, m) (typically called via std::allocator_traits<Alloc>::deallocate(a, result.ptr, m))
in order to avoid memory leak.
The argument m used in deallocation must be not less than n and not greater than result.count, otherwise, the behavior is undefined. Note that n is always equal
to result.count if the allocator does not provide allocate_at_least, which means that m is required to be equal to n.
| Feature-test macro | Value | Std | Comment |
|---|---|---|---|
| __cpp_lib_allocate_at_least | 202302L | (C++23) | allocate_at_least etc. |