Why double in C is 8 bytes aligned?

Question

I was reading a article about data types alignment in memory(here) and I am unable to understand one point i.e.

Note that a double variable will be allocated on 8 byte boundary on 32 bit machine and requires two memory read cycles. On a 64 bit machine, based on number of banks, double variable will be allocated on 8 byte boundary and requires only one memory read cycle.

My doubt is: Why double variables need to be allocated on 8 byte boundary and not on 4 byte? If it is allocated on 4 byte boundary still we need only 2 memory read cycles(on a 32 bit machine). Correct me if I am wrong.

Also if some one has a good tutorial on member/memory alignment, kindly share.

Answer 1

The reason to align a data value of size 2^N on a boundary of 2^N is to avoid the possibility that the value will be split across a cache line boundary.

The x86-32 processor can fetch a double from any word boundary (8 byte aligned or not) in at most two, 32-bit memory reads. But if the value is split across a cache line boundary, then the time to fetch the 2nd word may be quite long because of the need to fetch a 2nd cache line from memory. This produces poor processor performance unnecessarily. (As a practical matter, the current processors don't fetch 32-bits from the memory at a time; they tend to fetch much bigger values on much wider busses to enable really high data bandwidths; the actual time to fetch both words if they are in the same cache line, and already cached, may be just 1 clock).

A free consequence of this alignment scheme is that such values also do not cross page boundaries. This avoids the possibility of a page fault in the middle of an data fetch.

So, you should align doubles on 8 byte boundaries for performance reasons. And the compilers know this and just do it for you.