Is 8-byte alignment for “double” type necessary?

Question

I understand word-alignment, which makes the cpu only need to read once when reading an integer into a register.

But is 8-byte alignment (let\'s assume 32bit system

Answer 1

I just found the answer:

"6. When memory reading is efficient in reading 4 bytes at a time on 32 bit machine, why should a double type be aligned on 8 byte boundary?

It is important to note that most of the processors will have math co-processor, called Floating Point Unit (FPU). Any floating point operation in the code will be translated into FPU instructions. The main processor is nothing to do with floating point execution. All this will be done behind the scenes.

As per standard, double type will occupy 8 bytes. And, every floating point operation performed in FPU will be of 64 bit length. Even float types will be promoted to 64 bit prior to execution.

The 64 bit length of FPU registers forces double type to be allocated on 8 byte boundary. I am assuming (I don’t have concrete information) in case of FPU operations, data fetch might be different, I mean the data bus, since it goes to FPU. Hence, the address decoding will be different for double types (which is expected to be on 8 byte boundary). It means, the address decoding circuits of floating point unit will not have last 3 pins."