I worked with megafunctions to generate 32bit data memory in the fpga.but the output was addressed 32bit (4 bytes) at time , how to do 1 byte addressing ?
i have Altera Cyclone IV ep4ce6e22c8.
I'm designing a 32bit CPU in fpga ,
Nowadays every CPU address bus works in bytes. Thus to access your 32-bit wide memory you should NOT connect the LS 2 address bits. You can use the A[1:0] address bits to select a byte (or half word using A[1] only) from the memory when your read.
You still will need four byte write enable signals. This allows you to write word, half-words or bytes.
Have a look at existing CPU buses or existing connection standards like AHB or AXI.
Post edit:
but reading address 0001 , i get 0x05060708 but the desired value is 0x02030405.
What you are trying to do is read a word from a non-aligned address. There is no existing 32-bit wide memory that supports that. I suggest you have a look at how a 32-bit wide memory works.
The old Motorola 68020 architecture supported that. It requires a special memory controller which first reads the data from address 0 and then from address 4 and re-combines the data into a new 32-bit word.
With the cost of memory dropping and reducing CPU cycles becoming more important, no modern CPU supports that. They throw an exception: non-aligned memory access.
You have several choices:
Build a special memory controller which supports unaligned accesses.
Adjust your expectations.
I would go for the latter. In general it is based on the wrong idea how a memory works. As consolidation: You are not the first person on this website who thinks that is how you read words from memory.
Pagination in some processor make it possible to map virtual address
(A2345678) to physical address (823C5678). However, it is not possible
to map virtual address (345678) to (2ABC678). What can we conclude
about size of frame, page, size of virtual memory, size of physical
memory.
What I think about it:
(A2345678) -> (823C5678)
So, size offset is most 19 bits. We know that size of page (and frame) has size at most 219, like in my previous question.
When it comes to size of virtual memory, and physical memory - I can conclude nothing.
Similary, I don't know what tell me information about non-possibility mapping address.
Can you try to explain it me ?
I do see something we can conclude after all:
If a mapping to physical address 0x823C5678 is possible, physical memory is at least that large. (Assuming there aren't any holes in physical address space; not a good assumption on real hardware, but whatever. We can tell that physical address space is at least that big, even if it doesn't all map to DRAM or MMIO).
Similarly, the valid virtual address 0xA2345678 gives us a lower bound on the virtual address size. Presumably all the virtual address bits can be 1, so the highest possible virtual address is at least 0xFFFFFFFF. i.e. virtual addresses are at least 32 bits, but could be any larger size.
This reasoning applies to physical address space, but not the size of physical memory. (e.g. in a computer with 19GiB of RAM, the highest valid physical address isn't 2n-1.)
The fact that you can't map 0x345678 to 0x2ABC678 does tell us that the page size is greater than 212. The physical address is below the address that was mappable, so we can rule out that possible reason for the mapping being impossible. I think too high and misaligned are the only possible reasons for a mapping not being possible.
(0xc = 0b1100, while 0x5 is 0b0101, so the common bits are only 0x678.)
We can assume that physical memory is a whole number of pages, so we can round up the lowest possible end of physical memory to the next multiple of 213.
A microprocessor is byte addressable with 24bit address bus and 16bit data bus and one word contains two bytes. I was asked a question regarding attaching peripherals, adding memory, and address space and there's a few general concepts I don't see why they work.
Why is it that to calculate the address space you use the address bus not the data bus? Is the address space a function of the address bus or does it have to do with the microprocessor? How is it relevant that one word contains two bytes?
Why is it that to calculate the address space you use the address bus not the data bus?
Because it's the address bits that go out to the memory subsystem to tell them which memory location you want to read or write. The data bits just carry the data being read or written.
Is the address space a function of the address bus or does it have to do with the microprocessor?
Yes, the address space is a function of the address bus though there are tricks you can use to expand how much memory you can use.
An example of that is bank switching which gives you more accessible memory but no more address space (multiple blocks of memory co-exist at the same address, one at a time).
Another example is shown below where you can effectively double the usable memory, provided you're willing to only read and write words.
How is it relevant that one word contains two bytes?
The data bus size generally dictates the size of a memory cell. Larger memory cells can mean you can have more memory available to you but not more memory cells.
With your example, assuming you can only access words, you could get 16 megawords which is 32 megabytes.
This depends, of course, on how the memory is put together. It may be that you are able to access memory on individual byte boundaries (e.g., bytes 0/1 or 1/2 or 2/3) rather than just word boundaries, which would mean you don't actually get that full 32MB but only 16MB plus maybe one extra byte when you read the word at address FFFFFF).
I was reading up on Virtual Memory and from what I understand is that each process has its own VM table that maps VM addresses to Physical Addresses in real memory. So if a process allocated objects continuously they can potentially be stored in completely different places in Physical Memory. My question is that if I allocate and array which is supposed to be stored in a contiguous block of memory and if the size of the array requires more space than one page can provide, from what I understand is that array will be stored contiguously in VM but possibly in completely different location in PM. Is this correct? please correct me if I misunderstood how VM works. And if it is correct does that mean we are only concerned whether allocation is contiguous in VM?
Whether or not something that overlaps a page boundary is actually contiguous in Physical Memory is never really knowable with modern memory handlers. Memory glue logic essentially treats all addressable memory pages as an unordered set, and the ordering is essentially associated with a process; there's no guarantee that for different processes that end up getting assigned the same two physical memory pages (at different points in time) that the expressed relationship between those physical pages will be the same. Effectively, there's a translation layer between the CPU and the memory that handles this stuff.
That's right. Arrays must only looks contiguous for your application, but may be physically scattered on memory.
I just wanted to add/make it clear that from a user space program's point of view, a chunk of allocated memory always appears contiguous. The operating system in conjunction with the CPU's Memory Management Unit (MMU) handles all virtual to physical memory mappings and the programmer never needs to worry about how this mapping is handled (unless, of course, said programmer is writing an operating system).
A compiler (or one who writes code in assembly) can treat a program's addresses as starting from 0 and going up until the largest address needed for that particular program. The operating system then creates a page table for each process and uses this table to partially decode a physical address for each virtual memory location. The OS treats an address in a program as two separate parts, the page address and the offset into that page. Then, the MMU translates a page address into a physical frame address. Note that a physical memory "frame" is analogous to the conceptual "page" from the standpoint of the OS; these two are of the same size (eg 4096 bytes).
Since physical memory is divided into equally sized frames, and page size is the same as frame size you can know how much of your virtual address is used as a page location and how much is an offset into that page. For instance, if your OS "allocates" 4 gigabytes to each process (as is the case in Linux), and your page/frame size is 4096 bytes, you can know that 20 bits (4,294,967,296 bytes / 4096 bytes = 2 ^ 20 = 1,048,576 pages/page addresses) of a 32 bit address are used as a page address, which will then be converted to a physical frame address by the MMU, and the remaining 12 bits are used as an offset to determine the location of the address starting from the beginning of the page/frame.
VM (user pace) address --> page + offset (OS) --> frame + offset (MMU) = physical address
Admittedly I don't get it. Say you have a memory with a memory word of length of 1 byte. Why can't you access a 4 byte long variable in a single memory access on an unaligned address(i.e. not divisible by 4), as it's the case with aligned addresses?
The memory subsystem on a modern processor is restricted to accessing memory at the granularity and alignment of its word size; this is the case for a number of reasons.
Speed
Modern processors have multiple levels of cache memory that data must be pulled through; supporting single-byte reads would make the memory subsystem throughput tightly bound to the execution unit throughput (aka cpu-bound); this is all reminiscent of how PIO mode was surpassed by DMA for many of the same reasons in hard drives.
The CPU always reads at its word size (4 bytes on a 32-bit processor), so when you do an unaligned address access — on a processor that supports it — the processor is going to read multiple words. The CPU will read each word of memory that your requested address straddles. This causes an amplification of up to 2X the number of memory transactions required to access the requested data.
Because of this, it can very easily be slower to read two bytes than four. For example, say you have a struct in memory that looks like this:
struct mystruct {
char c; // one byte
int i; // four bytes
short s; // two bytes
}
On a 32-bit processor it would most likely be aligned like shown here:
The processor can read each of these members in one transaction.
Say you had a packed version of the struct, maybe from the network where it was packed for transmission efficiency; it might look something like this:
Reading the first byte is going to be the same.
When you ask the processor to give you 16 bits from 0x0005 it will have to read a word from 0x0004 and shift left 1 byte to place it in a 16-bit register; some extra work, but most can handle that in one cycle.
When you ask for 32 bits from 0x0001 you'll get a 2X amplification. The processor will read from 0x0000 into the result register and shift left 1 byte, then read again from 0x0004 into a temporary register, shift right 3 bytes, then OR it with the result register.
Range
For any given address space, if the architecture can assume that the 2 LSBs are always 0 (e.g., 32-bit machines) then it can access 4 times more memory (the 2 saved bits can represent 4 distinct states), or the same amount of memory with 2 bits for something like flags. Taking the 2 LSBs off of an address would give you a 4-byte alignment; also referred to as a stride of 4 bytes. Each time an address is incremented it is effectively incrementing bit 2, not bit 0, i.e., the last 2 bits will always continue to be 00.
This can even affect the physical design of the system. If the address bus needs 2 fewer bits, there can be 2 fewer pins on the CPU, and 2 fewer traces on the circuit board.
Atomicity
The CPU can operate on an aligned word of memory atomically, meaning that no other instruction can interrupt that operation. This is critical to the correct operation of many lock-free data structures and other concurrency paradigms.
Conclusion
The memory system of a processor is quite a bit more complex and involved than described here; a discussion on how an x86 processor actually addresses memory can help (many processors work similarly).
There are many more benefits to adhering to memory alignment that you can read at this IBM article.
A computer's primary use is to transform data. Modern memory architectures and technologies have been optimized over decades to facilitate getting more data, in, out, and between more and faster execution units–in a highly reliable way.
Bonus: Caches
Another alignment-for-performance that I alluded to previously is alignment on cache lines which are (for example, on some CPUs) 64B.
For more info on how much performance can be gained by leveraging caches, take a look at Gallery of Processor Cache Effects; from this question on cache-line sizes
Understanding of cache lines can be important for certain types of program optimizations. For example, the alignment of data may determine whether an operation touches one or two cache lines. As we saw in the example above, this can easily mean that in the misaligned case, the operation will be twice slower.
It's a limitation of many underlying processors. It can usually be worked around by doing 4 inefficient single byte fetches rather than one efficient word fetch, but many language specifiers decided it would be easier just to outlaw them and force everything to be aligned.
There is much more information in this link that the OP discovered.
you can with some processors (the nehalem can do this), but previously all memory access was aligned on a 64-bit (or 32-bit) line, because the bus is 64 bits wide, you had to fetch 64 bit at a time, and it was significantly easier to fetch these in aligned 'chunks' of 64 bits.
So, if you wanted to get a single byte, you fetched the 64-bit chunk and then masked off the bits you didn't want. Easy and fast if your byte was at the right end, but if it was in the middle of that 64-bit chunk, you'd have to mask off the unwanted bits and then shift the data over to the right place. Worse, if you wanted a 2 byte variable, but that was split across 2 chunks, then that required double the required memory accesses.
So, as everyone thinks memory is cheap, they just made the compiler align the data on the processor's chunk sizes so your code runs faster and more efficiently at the cost of wasted memory.
Fundamentally, the reason is because the memory bus has some specific length that is much, much smaller than the memory size.
So, the CPU reads out of the on-chip L1 cache, which is often 32KB these days. But the memory bus that connects the L1 cache to the CPU will have the vastly smaller width of the cache line size. This will be on the order of 128 bits.
So:
262,144 bits - size of memory
128 bits - size of bus
Misaligned accesses will occasionally overlap two cache lines, and this will require an entirely new cache read in order to obtain the data. It might even miss all the way out to the DRAM.
Furthermore, some part of the CPU will have to stand on its head to put together a single object out of these two different cache lines which each have a piece of the data. On one line, it will be in the very high order bits, in the other, the very low order bits.
There will be dedicated hardware fully integrated into the pipeline that handles moving aligned objects onto the necessary bits of the CPU data bus, but such hardware may be lacking for misaligned objects, because it probably makes more sense to use those transistors for speeding up correctly optimized programs.
In any case, the second memory read that is sometimes necessary would slow down the pipeline no matter how much special-purpose hardware was (hypothetically and foolishly) dedicated to patching up misaligned memory operations.
#joshperry has given an excellent answer to this question. In addition to his answer, I have some numbers that show graphically the effects which were described, especially the 2X amplification. Here's a link to a Google spreadsheet showing what the effect of different word alignments look like.
In addition here's a link to a Github gist with the code for the test.
The test code is adapted from the article written by Jonathan Rentzsch which #joshperry referenced. The tests were run on a Macbook Pro with a quad-core 2.8 GHz Intel Core i7 64-bit processor and 16GB of RAM.
If you have a 32bit data bus, the address bus address lines connected to the memory will start from A2, so only 32bit aligned addresses can be accessed in a single bus cycle.
So if a word spans an address alignment boundary - i.e. A0 for 16/32 bit data or A1 for 32 bit data are not zero, two bus cycles are required to obtain the data.
Some architectures/instruction sets do not support unaligned access and will generate an exception on such attempts, so compiler generated unaligned access code requires not just additional bus cycles, but additional instructions, making it even less efficient.
If a system with byte-addressable memory has a 32-bit-wide memory bus, that means there are effectively four byte-wide memory systems which are all wired to read or write the same address. An aligned 32-bit read will require information stored in the same address in all four memory systems, so all systems can supply data simultaneously. An unaligned 32-bit read would require some memory systems to return data from one address, and some to return data from the next higher address. Although there are some memory systems that are optimized to be able to fulfill such requests (in addition to their address, they effectively have a "plus one" signal which causes them to use an address one higher than specified) such a feature adds considerable cost and complexity to a memory system; most commodity memory systems simply cannot return portions of different 32-bit words at the same time.
On PowerPC you can load an integer from an odd address with no problems.
Sparc and I86 and (I think) Itatnium raise hardware exceptions when you try this.
One 32 bit load vs four 8 bit loads isnt going to make a lot of difference on most modern processors. Whether the data is already in cache or not will have a far greater effect.