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Showing content from https://en.cppreference.com/w/cpp/language/../memory/align.html below:

std::align - cppreference.com

Given a pointer ptr to a buffer of size space, returns a pointer aligned by the specified alignment for size number of bytes and decreases space argument by the number of bytes used for alignment. The first aligned address is returned.

The function modifies the pointer only if it would be possible to fit the wanted number of bytes aligned by the given alignment into the buffer. If the buffer is too small, the function does nothing and returns nullptr.

The behavior is undefined if alignment is not a power of two.

[edit] Parameters alignment - the desired alignment size - the size of the storage to be aligned ptr - pointer to contiguous storage (a buffer) of at least space bytes space - the size of the buffer in which to operate [edit] Return value

The adjusted value of ptr, or null pointer value if the space provided is too small.

[edit] Example

Demonstrates the use of std::align to place objects of different type in memory.

#include <iostream>
#include <memory>
#include <new>
 
template<std::size_t N>
struct MyAllocator
{
    std::byte data[N];
    std::size_t sz{N};
    void* p{data};
 
    MyAllocator() = default;
 
    // Note: only well-defined for implicit-lifetime types
    template<typename T>
    T* implicit_aligned_alloc(std::size_t a = alignof(T))
    {
        if (std::align(a, sizeof(T), p, sz))
        {
            T* result = std::launder(reinterpret_cast<T*>(p));
            p = static_cast<std::byte*>(p) + sizeof(T);
            sz -= sizeof(T);
            return result;
        }
        return nullptr;
    }
};
 
int main()
{
    MyAllocator<64> a;
    std::cout << "allocated a.data at " << (void*)a.data
              << " (" << sizeof a.data << " bytes)\n";
 
    // Allocate a char
    if (char* p = a.implicit_aligned_alloc<char>())
    {
        *p = 'a';
        std::cout << "allocated a char at " << (void*)p << '\n';
    }
 
    // Allocate an int
    if (int* p = a.implicit_aligned_alloc<int>())
    {
        *p = 1;
        std::cout << "allocated an int at " << (void*)p << '\n';
    }
 
    // Allocate an int, aligned at a 32-byte boundary
    if (int* p = a.implicit_aligned_alloc<int>(32))
    {
        *p = 2;
        std::cout << "allocated an int at " << (void*)p << " (32-byte alignment)\n";
    }
}

Possible output:

allocated a.data at 0x7ffc654e8530 (64 bytes)
allocated a char at 0x7ffc654e8530
allocated an int at 0x7ffc654e8534
allocated an int at 0x7ffc654e8540 (32-byte alignment)
[edit] Defect reports

The following behavior-changing defect reports were applied retroactively to previously published C++ standards.

DR Applied to Behavior as published Correct behavior LWG 2377 C++11 alignment required to be a fundamental or supported extended alignment value only need to be a power of two [edit] See also alignof (C++11) queries alignment requirements of a type
(operator)[edit] alignas (C++11) specifies that the storage for the variable should be aligned by specific amount
(specifier)[edit]

(since C++11)(deprecated in C++23)

defines the type suitable for use as uninitialized storage for types of given size
(class template) [edit] informs the compiler that a pointer is aligned
(function template) [edit]

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