How should #pragma pack(8) work?
<p>I'm new to structure alignment and packing. I thought I understood it, but I'm finding some results that I did not expect (see below).</p>

<p>My understanding of structure alignment is:</p>

<ol>
<li><p>Types are generally aligned on memory addresses that are multiples of their size.</p></li>
<li><p>Padding is added as needed to facilitate proper alignment</p></li>
<li><p>The end of a structure must be padded out to a multiple of the largest element (to facilitate array access)</p></li>
</ol>

<p>The <code>#pragma pack</code> directive basically allows overriding the general convention of aligning based on a type's size:</p>

<pre><code>#pragma pack(push, 8)
struct SPack8
{
  // Assume short int is 2 bytes, double is 8 bytes, and int is 4 bytes
  short int a;
  double b;
  short int c;
  int d;
};
#pragma pack(pop)

Pseudo struct layout: What I expected:
// note: PADDING IS BRACKETED
0, 1, [2, 3, 4, 5, 6, 7] // a occupies address 0, 1
8, 9, 10, 11, 12, 13, 14, 15, // b occupies 8-15 inclusive
16, 17, [18, 19, 20, 21, 22, 23] // c occupies 16-17 inclusive
24, 25, 26, 27 // d occupies 24-27 inclusive
// Thus far, SPack8 is 28 bytes, but the structure must be a multiple of
// sizeof(double) so we need to add padding to make it 32 bytes
[28, 29, 30, 31]
</code></pre>

<p>To my surprise, sizeof(SPack8) == 24 on VS 2015 x86. It seems that d is not being aligned on an 8 byte address:</p>

<pre><code>offsetof(SPack, a) // 0, as expected
offsetof(SPack, b) // 8, as expected
offsetof(Spack, c) // 16, as expected
offsetof(SPack, d) // 20..what??
</code></pre>

<p>Can someone please explain what is happening/what I have misunderstood?</p>

<p>Thank you!</p>