std::align(3)

NAME
std::align - std::align

Synopsis
   Defined in header <memory>
   void* align( std::size_t alignment,
   std::size_t size, (since C++11)
   void*& ptr,
   std::size_t& space );
   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.

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

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

Example
   demonstrates the use of std::align to place objects of different type in memory // Run this code
 #include <iostream>
 #include <memory>
 template <std::size_t N>
 struct MyAllocator
 {
     char data[N];
     void* p;
     std::size_t sz;
     MyAllocator() : p(data), sz(N) {}
     template <typename T>
     T* aligned_alloc(std::size_t a = alignof(T))
     {
         if (std::align(a, sizeof(T), p, sz))
         {
             T* result = reinterpret_cast<T*>(p);
             p = (char*)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.aligned_alloc<char>()) {
         *p = 'a';
         std::cout << "allocated a char at " << (void*)p << '\n';
     }
     // allocate an int
     if (int* p = a.aligned_alloc<int>()) {
         *p = 1;
         std::cout << "allocated an int at " << (void*)p << '\n';
     }
     // allocate an int, aligned at 32-byte boundary
     if (int* p = a.aligned_alloc<int>(32)) {
         *p = 2;
         std::cout << "allocated an int at " << (void*)p << " (32 byte alignment)\n";
     }
 }

Possible output:
 allocated a.data at 0x7ffd0b331f80 (64 bytes)
 allocated a char at 0x7ffd0b331f80
 allocated an int at 0x7ffd0b331f84
 allocated an int at 0x7ffd0b331fa0 (32 byte alignment)
  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 only need to be a power
                       or supported extended alignment value of two

See also
   alignof operator(C++11) queries alignment requirements of a type
   alignas specifier(C++11) specifies that the storage for the variable should be
                        aligned by specific amount
   aligned_storage defines the type suitable for use as uninitialized
   (C++11)(deprecated in C++23) storage for types of given size
                        (class template)
   assume_aligned informs the compiler that a pointer is aligned
   (C++20) (function template)