I am attempting to learn about memory alignment, without much success admittedly. I am using this article from IBM.
Can someone please explain to me what this excerpt means from the double byte memory access granularity section:
However, notice what happens when reading from address 1. Because the address doesn't fall evenly on the processor's memory access boundary, the processor has extra work to do. Such an address is known as an unaligned address. Because address 1 is unaligned, a processor with two-byte granularity must perform an extra memory access, slowing down the operation.
Why is another memory access in order? What does it mean by memory access boundary and it being even on the memory access boundary?
I have a VERY limited knowledge on the CPU, as I have only delt with upper level programming (Objective-C and C++). Any help is greatly appreciated!
Thanks!
The example is describing what happens when you try to read a block of 4 consecutive bytes on a CPU with double-byte access granuality. On this type of CPU, memory is accessed as pairs of bytes, always starting with an even-numbered byte.
If you try to read the block starting with byte 0, it has to perform 2 reads: bytes 0-1 and bytes 2-3.
If you try to read the block starting with byte 1, it has to perform 3 reads: bytes 0-1 (to get byte 1), bytes 2-3, and bytes 4-5 (to get byte 4).
Memory access granularity is the number of bytes it accesses at a time, and a memory access boundary is where each of these groups of bytes begins. The groups of bytes are always addressed at even multiples of the granularity -- if it's double-byte granularity they start on even addresses, if it's quad-byte granularity they're at multiples of 4.
As an analogy, consider an apartment building with 4 units on each floor. Units 0-3 are on floor 0, units 4-7 are on floor 1, etc. If you want to slip a flyer under the doors of units 0-3, you only have to go to one floor. But if you want to slip a flyer under 1-4, you have to go to 2 floors: floor 0 for 1-3, floor 2 for unit 4.
Related
I have encountered two pairs of concepts in Tannebaum's MOS:
word vs. byte addressability
aligned vs. non-aligned access
I don't really understand how these two pairs relate. In particular:
What would be the point of having byte addressable memory (1), with non-aligned access not allowed (2) ? It seems to me that (2) means we cannot take advantage of (1).
In other words, would:
word-addressable memory with non-aligned access allowed
offer to the programmer any possibilities that
word-addressable memory with non-aligned access not allowed
cannot?
Thank you
Word addressability:
In ye olde days there were computers that operated on 12, 14, and 36 bits. These were machine words. You could not address a byte in such systems—they didn't have bytes. You addressed words.
Aligned access:
It requires multiple cycles to access unaligned data. The processor has to take extra steps to grab the data in pieces. Some processors (notably RISC) prohibit unaligned accesses entirely for performance.
Such systems are still byte addressable because you can access individual bytes for byte operands.
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
Can someone explain what's the different between Word and Byte addressable? How is it related to memory size etc.?
A byte is a memory unit for storage
A memory chip is full of such bytes.
Memory units are addressable. That is the only way we can use memory.
In reality, memory is only byte addressable. It means:
A binary address always points to a single byte only.
A word is just a group of bytes – 2, 4, 8 depending upon the data bus size of the CPU.
To understand the memory operation fully, you must be familiar with the various registers of the CPU and the memory ports of the RAM. I assume you know their meaning:
MAR(memory address register)
MDR(memory data register)
PC(program counter register)
MBR(memory buffer register)
RAM has two kinds of memory ports:
32-bits for data/addresses
8-bit for OPCODE.
Suppose CPU wants to read a word (say 4 bytes) from the address xyz onwards. CPU would put the address on the MAR, sends a memory read signal to the memory controller chip. On receiving the address and read signal, memory controller would connect the data bus to 32-bit port and 4 bytes starting from the address xyz would flow out of the port to the MDR.
If the CPU wants to fetch the next instruction, it would put the address onto the PC register and sends a fetch signal to the memory controller. On receiving the address and fetch signal, memory controller would connect the data bus to 8-bit port and a single byte long opcode located at the address received would flow out of the RAM into the CPU's MDR.
So that is what it means when we say a certain register is memory addressable or byte addressable. Now what will happen when you put, say decimal 2 in binary on the MAR with an intention to read the word 2, not (byte no 2)?
Word no 2 means bytes 4, 5, 6, 7 for 32-bit machine. In real physical memory is byte addressable only. So there is a trick to handle word addressing.
When MAR is placed on the address bus, its 32-bits do not map onto the 32 address lines(0-31 respectively). Instead, MAR bit 0 is wired to address bus line 2, MAR bit 1 is wired to address bus line 3 and so on. The upper 2 bits of MAR are discarded since they are only needed for word addresses above 2^32 none of which are legal for our 32 bit machine.
Using this mapping, when MAR is 1, address 4 is put on the bus, when MAR is 2, address 8 is put on the bus and so forth.
It is a bit difficult in the beginning to understand. I learnt it from Andrew Tanenbaums's structured computer organisation.
This image should make it easy to understand:
http://i.stack.imgur.com/rpB7N.png
Simply put,
• In the byte addressing scheme, the first word starts at address 0, and
the second word starts at address 4.
• In the word addressing scheme, all bytes of the first word are located
in address 0, and all bytes of the second word are located in address 1.
The advantage of byte-addressability are clear when we consider applications that process data one byte at a time. Access of a single byte in a byte-addressable system requires only the issuing of a single address. In a 16–bit word addressable system, it is necessary first to compute the address of the word containing the byte, fetch that word, and then extract the byte from the two-byte word. Although the processes for byte extraction are well understood, they are less efficient than directly accessing the byte. For this reason, many modern machines are byte addressable.
Addressability is the size of a unit of memory that has its own address. It's also the smallest chunk of memory that you can modify without affecting its neighbours.
For example: a machine where bytes are the normal 8 bits, and the word-size = 4 bytes. If it's a word-addressable machine, there's no such thing as the address of the second byte of an int. Dealing with strings (e.g. an array like char str[]) becomes inconvenient, because you still store characters packed together. Modifying just str[1] means loading the word that contains it, doing some shift/and/or operations to apply the change, then doing a word store.
Note that this is different from a machine that doesn't allow unaligned word load/stores (where the low 2 bits of a word address have to be 0). Such machines usually have a byte load/store instruction. We're talking about machines without even that.
CPU addresses might actually still include the low bits, but require them to always be zero (or ignore them). However, after checking that they're zero, the could be discarded, so the rest of the memory system only sees the word address, where two adjacent words have an address that differs by 1 (not 4). However, on a 16-bit CPU where a register can only hold 64k different addresses, you wouldn't likely do this. Each separate CPU address would refer to a different 2 bytes of memory, instead of discarding the low bit. 2B word-addressable memory would let you address 128kiB of memory, instead of just 64kiB with byte-addressable memory.
Fun fact: ARM used to use the low 2 bits of an address as a shuffle control for unaligned word loads. (But it always had byte load/store instructions.)
See also:
https://en.wikipedia.org/wiki/Word-addressable
https://en.wikipedia.org/wiki/Byte_addressing
Note that bit-addressable memory could exist, but doesn't. 8-bit bytes are nearly universally standard now. (Ancient computers sometimes had larger bytes, see the history section of wikipedia's Byte article.)
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.