Question

According to MSDN, the /Zp command defaults to 8, which means 64-bit alignment boundaries are used. I have always assumed that for 32-bit applications, the MSVC compiler will use 32-bit boundaries. For example:

struct Test
{
   char foo;
   int bar;
};

The compiler will pad it like so:

struct Test
{
   char foo;
   char padding[3];
   int bar;
};

So, since /Zp8 is used by default, does that mean my padding becomes 7+4 bytes using the same example above:

struct Test
{
   char foo;
   char padding1[7];
   int bar;
   char padding2[4];
}; // Structure has 16 bytes, ending on an 8-byte boundary

This is a bit ridiculous isn’t it? Am I misunderstanding? Why is such a large padding used, it seems like a waste of space. Most types on a 32-bit system aren’t even going to use 64-bits, so the majority of variables would have padding (probably over 80%).

Answer 1

That’s not how it works. Members are aligned to a multiple of their size. Char to 1 byte, short to 2, int to 4, double to 8. The structure is padded at the end to ensure the members still align correctly when the struct is used in an array.

A packing of 8 means it stops trying to align members that are larger than 8. Which is a practical limit, the memory allocator doesn’t return addresses aligned better than 8. And double is brutally expensive if it isn’t aligned properly and ends up straddling a cache line. But otherwise a headache if you write SIMD code, it requires 16 byte alignment.