std::allocator<T>::deallocate

Deallocates the storage referenced by the pointer p, which must be a pointer obtained by an earlier call to allocate()or allocate_at_least()(since C++23).

# Declarations

void deallocate( T* p, std::size_t n );

(constexpr since C++20)

# Parameters

# Return value

(none)

# Example

#include <algorithm>
#include <cstddef>
#include <iostream>
#include <memory>
#include <string>
 
class S
{
    inline static int n{1};
    int m{};
    void pre() const { std::cout << "#" << m << std::string(m, ' '); }
public:
    S(int x) : m{n++} { pre(); std::cout << "S::S(" << x << ");\n"; }
    ~S() { pre(); std::cout << "S::~S();\n"; }
    void id() const { pre(); std::cout << "S::id();\n"; }
};
 
int main()
{
    constexpr std::size_t n{4};
    std::allocator<S> allocator;
    try
    {
        S* s = allocator.allocate(n); // may throw
        for (std::size_t i{}; i != n; ++i)
        {
        //  allocator.construct(&s[i], i + 42); // removed in C++20
            std::construct_at(&s[i], i + 42);   // since C++20
        }
        std::for_each_n(s, n, [](const auto& e) { e.id(); });
        std::destroy_n(s, n);
        allocator.deallocate(s, n);
    }
    catch (std::bad_array_new_length const& ex) { std::cout << ex.what() << '\n'; }
    catch (std::bad_alloc const& ex) { std::cout << ex.what() << '\n'; }
}

# See also