No it does not. The struct will be padded to the alignment requested, but it will not be aligned. There is a chance, however, that this will be allowed in C++17 (the fact that this C++17 proposal exists should be pretty good proof this can't work in C++11).
I have seen this appear to work with some memory allocators, but that was pure luck. For instance, some memory allocators will align their memory allocations to powers of 2 of the requested size (up to 4KB) as an optimization for the allocator (reduce memory fragmentation, possibly make it easier to reuse previously freed memory, etc...). However, the new/malloc implementations that are included in the OS X 10.7 and CentOS 6 systems that I tested do not do this and fail with the following code:
#include <stdlib.h>
#include <assert.h>
struct alignas(8) test_struct_8 { char data; };
struct alignas(16) test_struct_16 { char data; };
struct alignas(32) test_struct_32 { char data; };
struct alignas(64) test_struct_64 { char data; };
struct alignas(128) test_struct_128 { char data; };
struct alignas(256) test_struct_256 { char data; };
struct alignas(512) test_struct_512 { char data; };
int main() {
test_struct_8 *heap_8 = new test_struct_8;
test_struct_16 *heap_16 = new test_struct_16;
test_struct_32 *heap_32 = new test_struct_32;
test_struct_64 *heap_64 = new test_struct_64;
test_struct_128 *heap_128 = new test_struct_128;
test_struct_256 *heap_256 = new test_struct_256;
test_struct_512 *heap_512 = new test_struct_512;
#define IS_ALIGNED(addr,size) ((((size_t)(addr)) % (size)) == 0)
assert(IS_ALIGNED(heap_8, 8));
assert(IS_ALIGNED(heap_16, 16));
assert(IS_ALIGNED(heap_32, 32));
assert(IS_ALIGNED(heap_64, 64));
assert(IS_ALIGNED(heap_128, 128));
assert(IS_ALIGNED(heap_256, 256));
assert(IS_ALIGNED(heap_512, 512));
delete heap_8;
delete heap_16;
delete heap_32;
delete heap_64;
delete heap_128;
delete heap_256;
delete heap_512;
return 0;
}
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