Why is a double member in struct not aligned on 8 byte boundary?

Question

This is about memory alignment. In code below, I expected that the offset of b inside the structure to be 8 (32-bit machine). See here. There by, making b alway

Answer 1

According to the System V ABI for i386, page 28, double only gets 4 bytes alignment, but compilers are recommended to provide an option for 8 bytes as well. It appears this is what is implemented by GCC on Linux, the option being called -malign-double.

Another alternative is to use -m64 to get x86-64 object code, which is already the default on some systems including Mac OS X.

Answer 2

There is no guarantee about alignment, at all, in ANSI C.

The alignment happens with automatic variables more than with anything declared on the heap. If you're on a POSIX OS, use memalign(3) to receive memory that you're sure is aligned. Malloc may return memory at any offset. You can use compiler directives like __attribute__ ((__packed__)) to put in your own alignments.

Answer 3

I expected that the offset of b inside the structure to be 8 (32-bit machine). See here

Your references explain why it could be advantageous to 8-align doubles. This doesn't amount to a guarantee that doubles will always be 8-aligned. If your sources say that they're always 8-aligned, and you observe an implementation in which they aren't, then your sources are wrong.

From the GCC man page:

Aligning "double" variables on a two word boundary will produce code that runs somewhat faster on a Pentium at the expense of more memory.

So GCC's stated reason for 4-alignment is to save memory. You can use -malign-double and -mno-align-double to control that, although of course you risk creating binary incompatibilities if you share structs between code compiled with conflicting options. For particular objects/struct members you can use GCC's own __attribute__((aligned)), _Alignas (in C11) or alignas (in C++11), all of which can take an integer constant to specify the required alignment.