how does the processor read memory?

Question

I\'m trying to re-implement malloc and I need to understand the purpose of the alignment. As I understand it, if the memory is aligned, the code will be executed faster because

Answer 1

As I understand it, if the memory is aligned, the code will be executed faster because the processor won't have to take an extra step to recover the bits of memory that are cut.

It's not necessarily an execution thing, an x86 has variable length instructions starting with single 8 bit instructions on up to a handful to several bytes, its all about being unaligned. but they have taken measures to smooth that out for the most part.

If I have a 64 bit bus on the edge of my processor that doesn't mean edge of chip that means edge of the core. The other side of this is a memory controller that knows the bus protocol and is the first place the addresses start to be decoded and the transactions start to split up down other buses toward their destination.

It is very much architecture and bus design specific and you can have architectures with different buses over time or different versions you can get an arm with a 64 bus or a 32 bit bus for example. but lets say we have a not atypical situation where the bus is 64 bits wide and all transactions on that bus are aligned on a 64 bit boundary.

If I were to do a 64 bit write to 0x1000 that would be a single bus transaction, which these days is some sort of write address bus with some id x and a length of 0 (n-1) then the other side acks that I see you want to do a write with id x, I am ready to take your data. Then the processor uses the data bus with id x to send the data, one clock per 64 bits this is a single 64 bit so one clock on that bus. and maybe an ack comes back or maybe not.

But if I wanted to do a 64 bit write to 0x1004, what would happen is that turns into two transactions one complete 64 bit address/data transaction at address 0x1000 with only four byte lanes enabled lanes 4-7 (representing bytes at address 0x1004-0x1007). Then a complete transaction at 0x1008 with 4 byte lanes enabled, lanes 0-3. So the actual data movement across the bus goes from one clock to two, but there is also twice the overhead of the handshakes to get to those data cycles. On that bus it is very noticeable, how the overall system design is though you may feel it or not, or may have to do many of them to feel it or not. But the inefficiency is there, buried in the noise or not.

I think I understand that a 64-bit processor reads 64-bit by 64-bit memory.

Not a good assumption at all. 32 bit ARMs have 64 bit buses these days the ARMv6 and ARMv7s for example come with them or can.

Now, let's imagine that I have a structure with in order (without padding): a char, a short, a char, and an int. Why will the short be misaligned? We have all the data in the block! Why does it have to be on an address which is a multiple of 2. Same question for the integers and other types?

unsigned char a   0x1000
unsigned short b  0x1001
unsigned char c   0x1003
unsigned int d    0x1004

You would normally use the structure items in the code something.a something.b something.c something.d. When you access something.b that is a 16 bit transaction against the bus. In a 64 bit system you are correct that if aligned as I have addressed it, then the whole structure is being read when you do x = something.b but the processor is going to discard all but byte lanes 1 and 2 (discarding 0 and 3-7), then if you access something.c it will do another bus transaction at 0x1000 and discard all but lane 3.

When you do a write to something.b with a 64 bit bus only byte lanes 1 and 2 are enabled. Now where more pain comes in is if there is a cache it is likely also constructed of a 64 bit ram to mate up with this bus, doesn't have to, but lets assume it does. you want to write through the cache to something.b, a write transaction at 0x1000 with byte lanes 1 and 2 enabled 0, 3-7 disabled. The cache ultimately gets this transaction, it internally has to do a read-modify write because it is not a full 64 bit wide transaction (all lanes enabled) so you are taking hit with that read-modify write from a performance perspective as well. (same was true for the unaligned 64 bit write above).

The short is unaligned because when packed its address lsbit is set, to be aligned a 16 bit item in an 8 bit is a byte world needs to be zero, for a 32 bit item to be aligned the lower two bits of its address are zero, 64 bit, three zeros and so on.

Depending on the system you may end up on a 32 or 16 bit bus (not for memory so much these days) so you can end up with the multiple transfers thing.

Your highly efficient processors like MIPS and ARM took the approach of aligned instructions, and forced aligned transactions even in the something.b case that specifically doesn't have a penalty on a 32 nor 64 bit bus. The approach is performance over memory consumption, so the instructions are to some extent wasteful in their consumption to be more efficient in their fetching and execution. The data bus is likewise much simpler. When high level concepts like a struct in C are constructed there is memory waste in padding to align each item in the struct to gain performance.

unsigned char a   0x1000
unsigned short b  0x1002
unsigned char c   0x1004
unsigned int d    0x1008

as an example

I also have a second question: With the structure I mentioned before, how does the processor know when it reads its 64 bits that the first 8 bits correspond to a char, then the next 16 correspond to a short etc...?

unsigned char c   0x1003

the compiler generates a single byte sized read at address 0x1003, this turns in to that specific instruction with that address and the processor generates the bus transaction to do that, the other side of the processor bus then does its job and so on down the line.

The compiler in general does not turn a packed version of that struct into a single 64 bit transaction that gives you all of the items, you burn a 64 bit bus transaction for each item.

it is possible that depending on the instruction set, prefetcher, caches and so on that instead of using a struct at a high level you create a single 64 bit integer and you do the work in the code, then you might or might not gain performance. This is not expected to perform better on most architectures running with caches and such, but when you get into embedded systems where you may have some number of wait states on the ram or some number of wait states on the flash or whatever code storage there is you can find times where instead of fewer instructions and more data transactions you want more instructions and fewer data transactions. code is linear a code section like this read, mask and shift, mask and shift, etc. the instruction storage may have a burst mode for linear transactions but data transactions take as many clocks as they take.

A middle ground is to just make everything a 32 bit variable or a 64 bit, then it is all aligned and performs relatively well at the cost of more memory used.

Because folks don't understand alignment, have been spoiled by x86 programming, choose to use structs across compile domains (such a bad idea), the ARMs and others are tolerating unaligned accesses, you can very much feel the performance hit on those platforms as they are so efficient if everything is aligned, but when you do something unaligned it just generates more bus transactions making everything take longer. So the older arms would fault by default, the arm7 could have the fault disabled but would rotate the data around the word (nice trick for swapping 16 bit values in a word) rather than spill over into the next word, later architectures default to not fault on aligned or most folks set them to not fault on aligned and they read/write the unaligned transfers as one would hope/expect.

For every x86 chip you have in your computer you have several if not handfuls of non-x86 processors in that same computer or peripherals hanging off that computer (mouse, keyboard, monitor, etc). A lot of those are 8-bit 8051s and z80s, but also a lot of them are arm based. So there is lots of non-x86 development going on not just all the phones and tablets main processors. Those others desire to be low cost and low power so more efficiency in the coding both in its bus performance so the clock can be slower but also a balance of code/data usage overall to reduce the cost of the flash/ram.

It is quite difficult to force these alignment issues on an x86 platform there is a lot of overhead to overcome its architectural issues. But you can see this on more efficient platforms. Its like a train vs a sports car, something falls off a train a person jumps off or on there is so much momentum its not noticed one bit, but step change the mass on the sports car and you will feel it. So trying to do this on an x86 you are going to have to work a lot harder if you can even figure out how to do it. But on other platforms its easier to see the effects. Unless you find an 8086 chip and I suspect you can feel the differences there, would have to pull out my manual to confirm.

If you are lucky enough to have access to chip sources/simulations then you can see this kind of thing happening all over the place and can really start to hand tune your program (for that platform). Likewise you can see what caching, write buffering, instruction prefetching in its various forms and so on do for overall performance and at times create parallel periods of time where other not-so-efficient transactions can hide, and or intentional spare cycles are created so that transactions that take extra time can have a time slice.

Answer 2

The effects can even include correctness, not just performance: C Undefined Behaviour (UB) leading to possible segfaults or other misbehaviour if you have a short object that doesn't satisfy alignof(short). (Faulting is expected on ISAs where load/store instructions require alignment by default, like SPARC, and MIPS before MIPS64r6)

Or tearing of atomic operations if an _Atomic int doesn't have alignof(_Atomic int).

(Typically alignof(T) = sizeof(T) up to some size, often register width or wider, in any given ABI).

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malloc should return memory with alignof(max_align_t) because you don't have any type info about how the allocation will be used.

For allocations smaller than sizeof(max_align_t), you can return memory that's merely naturally aligned (e.g. a 4-byte allocation aligned by 4 bytes) if you want, because you know that storage can't be used for anything with a higher alignment requirement.

Over-aligned stuff like the dynamically-allocated equivalent of alignas (16) int32_t foo needs to use a special allocator like C11 aligned_alloc. If you're implementing your own allocator library, you probably want to support aligned_realloc and aligned_calloc, filling those gaps that ISO C leave for no apparent reason.

And make sure you don't implement the braindead ISO C++17 requirement for aligned_alloc to fail if the allocation size isn't a multiple of the alignment. Nobody wants an allocator that rejects an allocation of 101 floats starting on a 16-byte boundary, or much larger for better transparent hugepages. aligned_alloc function requirements and How to solve the 32-byte-alignment issue for AVX load/store operations?

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I think I understand that a 64-bit processor reads 64-bit by 64-bit memory

Nope. Data bus width and burst size, and load/store execution unit max width or actually-used width, don't have to be the same as width of integer registers, or however the CPU defines its bitness. (And in modern high performance CPUs typically aren't. e.g. 32-bit P5 Pentium had a 64-bit bus; modern 32-bit ARM has load/store-pair instructions that do atomic 64-bit accesses.)

Processors read whole cache lines from DRAM / L3 / L2 cache into L1d cache; 64 bytes on modern x86; 32 bytes on some other systems.

And when reading individual objects or array elements, they read from L1d cache with the element width. e.g. a uint16_t array may only benefit from alignment to a 2-byte boundary for 2-byte loads/stores.

Or if a compiler vectorizes a loop with SIMD, a uint16_t array can be read 16 or 32 bytes at a time, i.e. SIMD vectors of 8 or 16 elements. (Or even 64 with AVX512). Aligning arrays to the expected vector width can be helpful; unaligned SIMD load/store run fast on modern x86 when they don't cross a cache-line boundary.

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Cache-line splits and especially page-splits are where modern x86 slows down from misalignment; unaligned within a cache line generally not because they spend the transistors for fast unaligned load/store. Some other ISAs slow down, and some even fault, on any misalignment, even within a cache line. The solution is the same: give types natural alignment: alignof(T) = sizeof(T).

In your struct example, modern x86 CPUs will have no penalty even though the short is misaligned. alignof(int) = 4 in any normal ABI, so the whole struct has alignof(struct) = 4, so the char;short;char block starts at a 4-byte boundary. Thus the short is contained within a single 4-byte dword, not crossing any wider boundary. AMD and Intel both handle this with full efficiency. (And the x86 ISA guarantees that accesses to it are atomic, even uncached, on CPUs compatible with P5 Pentium or later: Why is integer assignment on a naturally aligned variable atomic on x86?)

Some non-x86 CPUs would have penalties for the misaligned short, or have to use other instructions. (Since you know the alignment relative to an aligned 32-bit chunk, for loads you'd probably do a 32-bit load and shift.)

So yes there's no problem accessing one single word containing the short, but the problem is for load-port hardware to extract and zero-extend (or sign-extend) that short into a full register. This is where x86 spends the transistors to make this fast. (@Eric's answer on a previous version of this question goes into more detail about the shifting required.)

Committing an unaligned store back into cache is also non-trivial. For example, L1d cache might have ECC (error-correction against bit flips) in 32-bit or 64-bit chunks (which I'll call "cache words"). Writing only part of a cache word is thus a problem for that reason, as well as for shifting it to an arbitrary byte boundary within the cache word you want to access. (Coalescing of adjacent narrow stores in the store buffer can produce a full-width commit that avoids an RMW cycle to update part of a word, in caches that handle narrow stores that way). Note that I'm saying "word" now because I'm talking about hardware that's more word-oriented instead of being designed around unaligned loads/stores the way modern x86 is. See Are there any modern CPUs where a cached byte store is actually slower than a word store? (storing a single byte is only slightly simpler than an unaligned short)

(If the short spans two cache words, it would of course needs to separate RMW cycles, one for each byte.)

And of course the short is misaligned for the simple reason that alignof(short) = 2 and it violates this ABI rule (assuming an ABI that does have that). So if you pass a pointer to it to some other function, you could get into trouble. Especially on CPUs that have fault-on-misaligned loads, instead of hardware handling that case when it turns out to be misaligned at runtime. Then you can get cases like Why does unaligned access to mmap'ed memory sometimes segfault on AMD64? where GCC auto-vectorization expected to reach a 16-byte boundary by doing some multiple of 2-byte elements scalar, so violating the ABI leads to a segfault on x86 (which is normally tolerant of misalignment.)

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For the full details on memory access, from DRAM RAS / CAS latency up to cache bandwidth and alignment, see What Every Programmer Should Know About Memory? It's pretty much still relevant / applicable

Also Purpose of memory alignment has a nice answer. There are plenty of other good answers in SO's memory-alignment tag.

For a more detailed look at (somewhat) modern Intel load/store execution units, see: https://electronics.stackexchange.com/questions/329789/how-can-cache-be-that-fast/329955#329955

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how does the processor know when it reads its 64 bits that the first 8 bits correspond to a char, then the next 16 correspond to a short etc...?

It doesn't, other than the fact it's running instructions which treat the data that way.

In asm / machine-code, everything is just bytes. Every instruction specifies exactly what to do with which data. It's up to the compiler (or human programmer) to implement variables with types, and the logic of a C program, on top of a raw array of bytes (main memory).

What I mean by that is that in asm, you can run any load or store instruction you want to, and it's up to you to use the right ones on the right addresses. You could load 4 bytes that overlap two adjacent int variable into a floating-point register, then and run addss (single-precision FP add) on it, and the CPU won't complain. But you probably don't want to because making the CPU interpret those 4 bytes as an IEEE754 binary32 float is unlikely to be meaningful.

Answer 3

modern processors and memory are built to optimize memory access as much as possible. One the current way of accessing memory is to address it not byte by byte but by an address of a bigger block, e.g. by an 8 byte blocks. You do not need 3 lower bits of the address this way. To access a certain byte within the block the processs needs to get the block at the aligned address, then shift and mask the byte. So, it gets slower.

When fields in the struct are not aligned, there is a risk of slowing down the access to them. Therefore, it is better to align them.

But the aslignment requirements are based on the underlying platform. For systems which support word access (32 bit), 4-byte alignment is ok, otherwise 8-byte can be used or some other. The compiler (and libc) knows the requirements.

So, in your example char, short, char, the short will start with an odd byte position if not padded. To access it, the system might eed to read the 64 bit word for the struct, then shift it 1 byte right and then mask 2 bytes in order to provide you with this byte.