How is memory alignment done usually?

Question

Have a look at these 2 cases:

class A {
    public:
    int a;

    A () { a = 10;}
    void foo () {std::cout << "a = " << a << std::endl;}
};

Here sizeof(A) gives 4 bytes, which makes sense.

class A {
    public:
    int a;

    A () { a = 10;}
    virtual void foo () {std::cout << "a = " << a << std::endl;}
};

Here sizeof(A) gives 16 bytes as opposed to 12 bytes (4 + 8 for pointer).

Is there any explanation in terms of memory alignment for this ?

@shrinidhisondur: Because a 4 byte `int` is happy sitting on a 4 byte boundary. An 8 byte pointer, on the other hand, wants an 8 byte boundary of its own. As Tony points out, in order for the v-table pointer in the second array element to also be on an 8 byte boundary, `sizeof (A)` needs to be a multiple of 8. — Ben Voigt, Oct 29 '14 at 02:56
see also [this thread](http://stackoverflow.com/questions/9439240/sizeof-class-with-int-function-virtual-function-in-c) — M.M, Oct 29 '14 at 03:22

score 1 · Answer 1 · answered Oct 29 '14 at 01:41

It's up to the compiler how virtual functions are implemented, but what's likely happening here is it wants/needs to align the 8-byte pointer to the virtual dispatch table on a multiple-of-8 memory address. Then there's either { 4 bytes a, 4 padding, 8 vdt pointer } or { 8 vdt pointer, 4 bytes a, 4 padding } - the latter's less obvious, but consider that arrays of A need to be contiguous and spaced per sizeof(A), so 12's rounded up to 16 given the 8-byte alignment.

bigbigtree · Answer 2 · 2014-10-29T12:56:34.917

FIRST in virtual dispatch table, visual functions need a pointer size , in 32 bits computer, the size of pointer is 4 bytes, int 64 bits it's 8 bytes. so I think your computer is 64 bits.

SECOND the sizeof(A) should consider the padding and memory alignment . so 16 bytes is arranged like that: 4 bytes(int a) + 8 bytes(a virtual function pointer int 64 bits) + 4 bytes padding(because the max size of elements in A is the virtual function pointer which is 8 bytes, so the sizeof(A)should be the integral multiples of 8 bytes )

How is memory alignment done usually?

2 Answers2