So the architecture doesnt really matter other than x86 has traditionally not directly told you not to where mips and arm traditionally generate a data abort rather than trying to just make it work.
where it doesnt matter is that all processors have a fixed number of pins a fixed size (maximum) data bus a fixed size (max) address bus, "modern processors" tend to have data busses more than 8 bits wide but the units on addresses is still an 8 bit byte, so the opportunity for unaligned exists. Anything larger than one byte in a particular transfer has the opportunity of being unaligned if the architecture allows.
Transfers are typically in some units of bytes and/or bus widths. On an ARM amba/axi bus for example the length field is in units of bus widths, 32 or 64 bits, 4 or 8 bytes. And no it is not going to be in units of 4Kbytes....
(yes this is elementary I assume you understand all of this).
Whether it is 16 bits or 128 bits, the penalty for unaligned comes from the additional bus cycles which these days is an extra bus clock per. So for an ARM 16 bit unaligned transfer (which arm will support on its newer cores without faulting) that means you need to read 128 bits instead of 64, 64 bits to get 16 is not a penalty as 64 is the smallest size for a bus transfer. Each transfer whether it is a single width of the data bus or multiple has multiple clock cycles associated with it, lets say there are 6 clock cycles to do an aligned 16 bit read, then ideally it is 7 cycles to do an unaligned 16 bit. Seems small but it does add up.
caches help alot because the dram side of the cache will be setup to use multiples of the bus width and will always do aligned accesses for cache fetches and evictions. not-cached accesses will follow the same pain except the dram side is not handfuls of clocks but dozens to hundreds of clocks of overhead.
For random access a single 16 bit read that not only spans a bus width boundary but also happens to cross a cache line boundary will not just incur the one additional clock on the processor side but worst case it can incur an addition cache line fetch which is dozens to hundreds of additional clock cycles. if you were walking through an array of things that happen to not be aligned (structures/unions may be an example depending on the compiler and code) that additional cache line fetch would have happened anyway, if the array of things is a little over on one or both ends then you might still incur one or two more cache line fetches that you would have avoided had the array been aligned.
That is really the key to this on reads is before or after an aligned area you might have to incur a transfer for each one for each side you spill into.
Writes are both good and bad. random reads are slower because the transaction has to stall until the answer comes back. For a random write the memory controller has all the information it needs it has the address, data, byte mask, transfer type, etc. So it is fire and forget the processor has done its job and can call the transaction complete from its perspective and move on. Naturally gang too much of these up or do a read on something just written and then the processor stalls due to the completion of a prior write in addition to the current transaction.
An unaligned 16 bit write for example does not only incur the additional read cycle but assuming a 32 or 64 bit wide bus that would be one byte per location so you have to do a read-modify-write on whatever that closest memory is (cache or dram). so depending on how the processor and then memory controller implements it it can be two individual read-modify-write transactions (unlikely since that incurs twice the overhead), or the double width read, modify both parts, and a double width read. incurring two additional clocks over and above the overhead, the overhead is doubled as well. If it had been an aligned bus width write then no read-modify-write is required, you save the read. Now if this read-modify-write is in the cache then that is pretty fast but still noticeable up to a few clocks depending on what is queued up and you have to wait on.
I am also most familiar with ARM. Arm traditionally would punish an unaligned access with an abort, you could turn that off, and you would instead get a rotation of the bus rather than it spilling over which would make for some nice freebie endian swaps. the more modern arm cores will tolerate and implement an unaligned transfer. Understand for example a store multiple of say 4 or more registers against a non-64-bit-aligned address is not considered an unaligned access even though it is a 128 bit write to an address that is neither 64 nor 128 bit aligned. What the processor does in that case is brakes it into 3 writes, an aligned 32 bit write, an aligned 64 bit write and an aligned 32 bit write. the memory controller does not have to deal with the unaligned stuff. That is for legal things like store multiple. the core I am familiar with wont do a write length of more than 2 anyway, an 8 register store multiple, is not a single length of 4 write it is 2 separate length of two writes. But a load multiple of 8 registers, so long it is aligned on a 64 bit address is a single length of 4 transaction. I am pretty sure that since there is no masking on the bus side for a read, everything is in units of bus width, there is no reason to break say a 4 register load multiple on an address that is not 64 bits aligned into 3 transactions, simply do a length of 3 read. When the processor reads a single byte you cant tell that from the bus all you see is a 64 bit read AFAIK. The processor strips the byte lane out. If the processor/bus does care be it arm, x86, mips, etc, then sure you will hopefully see separate transfers.
Does everyone do this? no older processors (not thinking of an arm nor x86) would put more burden on the memory controller. I dont know what modern x86 and mips and such do.
Your malloc example. First off you are not going to see single bus transfers of 4Kbytes, that 4k will be broken up into digestible bits anyway. first off it has to do one to many bus cycles against the memory management unit to find the physical address and other properties anyway (those answers can get cached to make them faster, but sometimes they have to go all the way out to slow dram) so for that example the only transfer that matters is an aligned transfer that splits the 4k boundary, say a 16 bit transfer, for the mmu system to work at all the only way for that to be supported is that has to be turned into two separate 8 bit transfers that happen in those physical address spaces, and yes that literally doubles everything the mmu lookup cycles the cache/dram bus cycles, etc. Other than that boundary there is nothing special about your 8k being split. the bulk of your cycles will be within one of the two 4k pages, so it looks like any other random access, with of course repetitive/sequential accesses gaining the benefit of caching.
The short answer is that no matter what platform you are on either 1) the platform will abort an unaligned transfer, or 2) somewhere in the path there is an additional one or more (dozens/hundreds) as a result of the unaligned access compared to an aligned access.
https://stackoverflow.com/questions/23593994
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