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.
I need help understanding memory. How many bytes can be stored in a memory unit that uses 8 address bits and a 16 bit architecture?
I think it's 2^8 = 256. Is this correct?
Edit: I mean 256
It depends.
Firstly "16 bit architecture" is too vague to be a helpful characterization for this problem. A characterization like that generally refers to the width of registers and data paths (e.g. in the ALU), not how memory is addressed.
Secondly, the answer actually depends on whether the addresses are byte addresses or "word" addresses. AFAIK "almost all" new processor / instruction set architectures designed since the 1980's have used byte addresses. But prior to that it was common for addresses to address words of up to 60 bits (or possibly more).
But assuming byte addressing, then an 8 bit address allows you to address 2^8 bytes; i.e. 256 bytes.
On the other hand, if we assume word addressing with a 16 bit word, then 8 bit addresses will address 256 words ... which is 512 bytes.
The base answer is the 2^number of bits. However,long ago sixteen bit systems came up with means for accessing more than 2^16 memory though segments. While an application can only access 2^16 bytes at a time, changing the values in hardware registers allows the application to change which 2^16 bytes of a larger address space are being accessed.
You typically do things like
Map buffer 1 to the address space.
Queue a read operation to the buffer.
Map another buffer 2 to the address space.
Read operation completes-map Buffer 1 back so that the data can be accessed.
Endianess determines the ordering of bytes in a word. Let us consider the following
memory system :
so this is a byte addressable 32 bit memory.
If I move a hex value 'val = 0x56789A' into the memory location with word address 0, it will look like this for big endian:
And like this for little endian:
but we know that in the register the values are stored as '56789A' itslef, so that is no problem if we have a big endian type byte ordering, as the value can be loaded in the correct order.
but what about in the case of little endian where the order has to be reversed? In little endian it will be loaded as '9A7856', which is wrong. Then do we store it like this in the memory, (i.e.) have a different organization for little endian type ordering? such as :
Now we can load the value at word address location '0' into a register for any further operation.
this was a possible solution.
but now it would mean that the endianess will dependend on the memory cell arrangement and not exactly the processor...... How exactly does this work?
Or is it the case that endianess is not at all affected by the memory architecture but rather ONLY the processor? So finally Does endianess depend on processor or memory?
Endianness is really a question of interfaces. For a 32 bit register load one interface is asking for a 32 bit value, and the memory interface provides an array of bytes. Something has to resolve that interface. If it resolves the byte array as the low order bytes going into the most significant bits of the returned value, then it is big endian. If it resolves it with the low order bytes going into the least significant bits of the returned value it is little endian.
So, really your question is who resolves those interfaces. Technically, it matters how the processor requests data from the from the memory. If it requests it by saying "I want a 32 bit value" then the memory, which has an array of bytes, must resolve it. If it requests it by saying "I want 4 bytes", then the processor has to resolve it before storing it into the registers.
By convention, the processor resolves this interface. This allows the memory to work with both big and little endian processors because it can present the same interface (a byte array) to both types of processors and just let the processor resolve it anyway it wants it. At the risk of being to simplistic: the processor resolves it in the load/store unit and memory interface unit. The registers never have to think about it because by the time it reaches them, it is an 32 bit value, having already been resolved by the memory interface unit from the byte array it requested from memory.
Assume 32 Bit OS.
One memory location in a computer stores how much data?
Whats the basic unit of memory storage in a computer?
For Example to a store a integer what will be the memory addresses required?
If basic unit is BYTE the integer requires 4 bytes.
So if I need to store a byte then if start putting in the 1st byte in memory location
0001 then will my integer end at 0003 memory location?
Please correct me if am wrong?
Most commonly, modern systems are what you call "byte-accessible".
This means:
One memory location stores 1 byte (8 bits).
The basic storage unit for memory is 1 byte.
If you need to store 4 bytes, and place the first byte at 0001, the last byte will be at 0004. That's one byte at each of 0001, 0002, 0003, and 0004.
Keep in mind while systems have different CPU word sizes (a 32-bit system has a 32-bit or 4-byte word), memory is usually addressed by byte. The CPU's registers used in arithmetic are 4 bytes, but the "memory" programmers use for data storage is addressed in bytes.
On x86 systems, many memory-accessing instructions require values in memory to be "aligned" to addresses evenly divisible by the word size. e.g. 0x???0, 0x???4, 0x???8, 0x???C. So, storing an int at 0001 won't happen on most systems. Non-numeric data types can usually be found at any address.
See Wikipedia: Alignment Word (Computing) Memory Address
One memory location in a computer stores how much data?
It depends on the computer. A memory location means a part of memory that the CPU can address directly.
Whats the basic unit of memory storage in a computer?
It is the Bit, and then the Byte, but different CPUs are more comfortable addressing memory in words of particular sizes.
For Example to a store a integer what will be the memory addresses required? If basic unit is BYTE the integer requires 4 bytes.
In mathematics, the integer numbers are infinite, so infinite memory should be required to represent all/any of them. The choice made by a computer architecture about how much memory should be used to represent an integer is arbitrary. In the end, the logic about how integers are represented and manipulated is in software, even if it is embedded in the firmware. The programming language Python has an unbounded representation for integers (but please don't try a googol on it).
In the end, all computer architectures somehow allow addressing down to the Byte or Bit level, but they work best with addresses at their word size, which generally matches the bit-size of the CPU registers.
It is not about the amount of data, or the size of integers, but about the number of memory addresses the computer can use.
There are 4GiB addresses (for bytes) in 32 bits. To manage a cluster of machines with more than 4GiB of RAM, each system must manage larger addresses.
Again, it is all about the addressable memory space, and not about the size of integers. There were 64 bit integers even when CPUs preferred 8bit word addressing.
Depends on the architecture. 32-bits for 32-bits. 64-bits for 64-bits.
Usually it's called a "word"
Most values need to be aligned, so the addresses end with 0 4 8 or C
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.