std::align
From cppreference.com
| Defined in header <memory>
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| void* align( std::size_t alignment, std::size_t size, |
(since C++11) | |
If it is possible to fit size bytes of storage aligned by alignment into the buffer pointed to by ptr with length space, the function modifies ptr to point to the first possible address of such aligned storage and decreases space by the number of bytes used for alignment. If it is impossible (the buffer is too small), align does nothing.
Contents |
Parameters
| alignment | - | the desired alignment |
| size | - | the size of the storage to be aligned |
| ptr | - | pointer to contiguous storage of at least space bytes
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| 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); return result; } return nullptr; } }; int main() { MyAllocator<64> a; // allocate a char char* p1 = a.aligned_alloc<char>(); if (p1) *p1 = 'a'; std::cout << "allocated a char at " << (void*)p1 << '\n'; // allocate an int int* p2 = a.aligned_alloc<int>(); if (p2) *p2 = 1; std::cout << "allocated an int at " << (void*)p2 << '\n'; // allocate an int, aligned at 32-byte boundary int* p3 = a.aligned_alloc<int>(32); if (p3) *p3 = 2; std::cout << "allocated an int at " << (void*)p3 << " (32 byte alignment)\n"; }
Possible output:
allocated a char at 0x2ff21a08 allocated an int at 0x2ff21a0c allocated an int at 0x2ff21a20 (32 byte alignment)
See also
| (C++11)(deprecated in C++23) |
defines the type suitable for use as uninitialized storage for types of given size (class template) |
