std::allocate_at_least

template< class Alloc >
[[nodiscard]] constexpr std::allocation_result<
    typename std::allocator_traits<Alloc>::pointer>
    allocate_at_least( Alloc& a, std::size_t n );

std::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};.

For an Allocator a, std::allocator_at_least tries to allocate a storage for at least n Alloc::value_type objects, and provides a fallback mechanism (like those provided by std::allocator_traits) that allocates a storage for exact n objects.

Parameters

Return value

a.allocate_at_least(n) if it is well-formed, otherwise, {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 std::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::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 support allocate_at_least, which means that m is required to be equal to n.

Example

See also