std::align

Header: <memory>

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.

# Declarations

void* align( std::size_t alignment,
std::size_t size,
void*& ptr,
std::size_t& space );

(since C++11)

# Parameters

# Return value

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

# Example

#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";
    }
}

# Defect reports

DRApplied toBehavior as publishedCorrect behavior
LWG 2377C++11alignment required to be a fundamental or supported extended alignment valueonly need to be a power of two

# See also