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.
Related
So i did find some answers related to this in stackvoerflow but non of them clearly answered this
so if our memory is byte addressable and the word size is for example 4 byte, then why not make the memory byte addressable?
if i'm not mistaking CPU will work with words right? so when the cpu tries to get a word from the memory what's the difference between getting a 4 byte word from a byte addressable memory vs getting a word from word addressable memory?
if i'm not mistaking CPU will work with words right?
It depends on the Instruction Set Architecture (ISA) implemented by the CPU. For example, x86 supports operands of sizes ranging from a single 8-bit byte to as much as 64 bytes (in the most recent CPUs). Although the word size in modern x86 CPUs is 8 or 4 bytes only. The word size is generally defined as equal to the size of a general-purpose register. However, the granularity of accessing memory or registers is not necessarily restricted to the word size. This is very convenient from a programmer's perspective and from the CPU implementation perspective as I'll discuss next.
so when the cpu tries to get a word from the memory what's the
difference between getting a 4 byte word from a byte addressable
memory vs getting a word from word addressable memory?
While an ISA may support byte addressability, a CPU that implements the ISA may not necessarily fetch data from memory one byte at a time. Spatial locality of reference is a memory access pattern very common in most real programs. If the CPU was to issue single-byte requests along the memory hierarchy, it would unnecessarily consume a lot of energy and significantly hurt performance to handle single-byte requests and move one-byte data across the hierarchy. Therefore, typically, when the CPU issues a memory request for data of some size at some address, a whole block of memory (known as a cache line, which is usually 64-byte in size and 64-byte aligned) is brought to the L1 cache. All requests to the same cache line can be effectively combined into a single request. Therefore, the address bus between different levels of the memory hierarchy does not have to include wires for the bits that constitute an offset within the cache line. In that case, the implementation would be really addressing memory at the 64-byte granularity.
It can be useful, however, to support byte addressability in the implementation . For example, if only one byte of a cache line has changed and the cache line has to be written back to main memory, instead of sending all the 64 bytes to memory, it would take less energy, bandwidth, and time to send only the byte that changed (or few bytes). Another situation where byte addressability is useful is when providing support for the idea of critical-word first. This is much more to it, but to keep the answer simple, I'll stop here.
DDR SDRAM is a prevalent class of main memory interfaces used in most computer systems today. The data bus width is 8 bytes in size and the protocol supports only transferring aligned 8-byte chunks with byte enable signals (called data masks) to select which bytes to write. Therefore, main memory is typically 8-byte addressable. It is the CPU that provides the illusion of byte addressability.
memory normally is byte-addressable. But whole-word loads are possible, and get 4x as much data in the same time.
There's basically no difference, if the word load is naturally aligned; the low bits of the address are zero instead of being not present.
I am fairly new to programming and am starting to learn the ins and outs of memory allocation. One question that recently occurred to me that I haven't yet been able to find a clear answer to is do memory addresses themselves take up memory. For example, in a 32-bit system, the way I understand it is that each address in 4 bytes and they will typically refer to an empty 'bucket' in memory that is capable of storing 1 byte of data. Does this mean that for each memory location in a 32-bit system, we are actually using 5 bytes of memory(meaning 4 for the address and 1 for the empty bucket)? I'm sure I am missing something here but any clarification would be much appreciated. Thanks!
To reference a memory address you need to express that memory address somehow, and on 32 bit system a memory reference takes 4 bytes indeed. So for any addressable memory address, somewhere else in memory there are 4 bytes that have that address.
But this does not cascade to an x5 multiplication, because a program does not need to reference every byte of memory. It only needs the address where something in memory starts, and then it can work its way to every byte of that 'something' using arithmetic.
To give an example: you have a string in memory Justin Foss. Is it at address 0x10000000, and this address is stored in a variable. So the actual variable value is 0x10000000, pointing to the string Justin Foss. But at 0x10000000 you only have one byte, the J. At 0x10000001 there is the u, at 0x10000002 is the s and so on. Your application does not need a variable for each character, it only needs one variable (4 bytes) to the beginning of the string. Same for object (fields): you only store the address where the objects starts, and the compiler know how to do the arithmetic to find the field it needs, by adding the necessary offset. In general memory objects are quite large, and a few 4 byte variables in the program reference quite a bit of memory.
(at the risk of oversimplification) Memory is sequential. Address 123 is the one-hundred and twenty third byte from the first (zeroth) by in the system. There is no memory devoted to indicating byte 123 is 123. The byte that comes after that is 124.
I'm looking into reading single bits from memory (RAM, harddisk). My understanding was, one can not read less than a byte.
However I read someone telling it can be done with assembly.
I wan't the bandwidth usage to be as low as possible and the to be retrieved data is not sequential, so I can not read a byte and convert it to 8 bits.
I don't think the CPU will read less than the size of a cache line from RAM (64 bytes on recent Intel chips). From disk, the minimum is typically 4 kiB.
Reading a single bit at a time is neither possible nor necessary, since the data bus is much wider than that.
You cannot read less than a byte from any PC or hard disk that I know of. Even if you could, it would be extremely inefficient.
Some machines do memory mapped port io that can read/write less than a byte to the port, but it still shows up when you get it as at least a byte.
Use the bitwise operators to pick off specific bits as in:
char someByte = 0x3D; // In binary, 111101
bool flag = someByte & 1; // Get the first bit, 1
flag = someByte & 2; // Get the second bit, 0
// And so on. The number after the & operator is a power of 2 if you want to isolate one bit.
// You can also pick off several bits like so:
int value = someByte & 3; // Assume the lower 2 bits are interesting for some reason
It used to be, say 386/486 days, where a memory was a bit wide, 1 meg by 1 bit, but you will have 8 or some multiple number of chips, one for each bit lane on the bus, and you could only read in widths of the bus. today the memories are a byte wide and you can only read in units of 32 or 64 or multiples of those. Even when you read a byte, most designs fill in the whole byte. it adds unnecessarily complication/cost, to isolate the bus all the way to the memory, a byte read looks to most of the system as a 32 or 64 bit read, as it approaches the edge of the processor (sometimes physical pins, sometimes the edge of the core inside the chip) is when the individual byte lane is separated out and the other bits are discarded. Having the cache on changes the smallest divisible read size from the memory, you will see a burst or block of reads.
it is possible to design a memory system that is 8 bits wide and read 8 bits at a time, but why would you? unless it is an 8 bit processor which you probably couldnt take advantage of a 8bit by 2 gig memory. dram is pretty slow anyway, something like 133 mhz (even your 1600mhz memory is only short burst as you read from slow parts, memory has not gotten faster in over 10 years).
Hard disks are similar but different, I think sectors are the smallest divisible unit, you have to read or write in those units. so when reading you have a memory cycle on the processor, no different that going to a memory, and depending on the controller either before you do the read or as a result, a sector is read of the disk, into a buffer, not unlike a cache line read, then your memory cycle to the buffer in the disk controller either causes a bus width read and the processor divides it up or if the bus adds complexity to isolate byte lanes then you isolate a byte, but nobody isolates bit lanes. (I say the word nobody and someone will come back with an exception...)
most of this is well documented, not hard to find. For arm platforms look for the amba and/or axi specifications, freely downloaded. the number of bridges, pcie controllers, disk controller documents are all available for PCs and other platforms. it still boils down to an address and data bus or one goesouta and one goesinta data bus and some control signals that indicate the access type. some busses have byte lane enables, which is generally for a write not a read. If I want to write only a byte to a dram in a modern 64 bit system, I DO have to tell everyone almost all the way out to the dram what I want to write. To write a byte on a memory module which must be accessed 64 bits at a time, at a minimum a 64 bit read happens into a temporary place either the cache or the memory controller, then the byte to be written modifies the specific byte within the 64 bit word, then that 64 bit quantity, eventually, is written back to the memory module itself. You can do this using a combination of the address bits and a few control signals or you can just put 8 byte lane enables and the lower address bits can be ignored. Hard disk, same deal, have to read a sector, modify one byte, then eventually write the whole sector at a time. with flash and eeprom, you can only write zeros (from the programmers perspective), you erase to ones (from the programmers perspective, is actually a zero in the logic, there is an inversion) and a write has to be a sector at a time, sectors can be 64 bytes, 128 bytes, 256 bytes typically.
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.