std::align

 

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

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 or supported extended alignment value	only need to be a power 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 (C++11)(deprecated in C++23)	defines the type suitable for use as uninitialized storage for types of given size (class template)
assume_aligned (C++20)	informs the compiler that a pointer is aligned (function template)