Process segments(stack,heap,data and code) are contiguous in memory? - process

Can some body help, does all segments belongs to process are contiguous in the memory? if yes then why those should be in contiguous?.
as per my understanding stack comes down and heap grows up(depends on CPU architecture) that shows all segments lay one by one(tightly coupled).

Segments themselves are architecturally dependent. Few processors other than Intel use segments. In addition, there are no segments in 64-bit mode on Intel processors.
If you are on an intel processor, not-64-bit-mode, the segments are virtually contiguous. The heap does not interleaves with the code. However, their may be discontinuities in the virtual memory within a segment in systems that support that.The segments do not have to be contiguous with each other.
If you are running on a system without segments, the code, heap, and data can be interleaves. However, each stack must be contiguous.

On x86 in virtual memory yes they are contiguous from processes point of view but the portions of these segments can be mapped to any random physical memory portions.

Related

Cortex-M3 External RAM Region

I'm currently researching topics such as RAM/ROM/Stack/Heap and data segments etc.
I was looking at the ARM Cortex-M3 memory map and saw the region labeled "External RAM".
According to the data sheet of a random Cortex-M3 STM32 MCU the external RAM region is mapped from 0x60000000- 0x9FFFFFFF, so it is quite large!
I couldn't find a definitive answer about how this region is actually used.
I imagine you would have an external SRAM and you would choose between two options.
(1) Read via the SPI interface and place into a local buffer(stack), then load that local buffer into the external ram region. This option seems to have a lot of negative consequences, such as hogging the CPU and increasing the stack temporarily if the requested data is very large.
(2) Utilize a DMA and transfer from the SPI interface into the external ram region.
Now I can't understand, why you would map the data to this specific address range, what are the advantages, why don't you just place the data directly in that huge memory region?
Now I'm asking this question because I have a slight feeling I have completely missed the point of what the External RAM region really is.
-Edit-
In the data sheet that is linking to the STM32 device, the memory region "External RAM" is marked as reserved. It is my conclusion that the memory regions listed by ARM is showing the full potential of a 32bit MCU, as I incorrectly state that the external RAM region "is quite large!" does not necessarily mean that this is "real" size of that region, if it is even used, it depends on what the vendor can physically achieve within the MCU hardware, and I imagine they would limit hardware capabilities to be competitive on price, power consumption etc.
I imagine you would have an external [SRAM][3] and you would choose
between two options.
(1) Read via the SPI interface and place into a local buffer(stack), then load that local buffer into the external ram region. This option
seems to have a lot of negative consequences, such as hogging the CPU
and increasing the stack temporarily if the requested data is very
large.
(2) Utilize a DMA and transfer from the SPI interface into the external ram region.
None of the above. External memory on an SPI bus is not memory mapped. If you have an SPI memory, it is not mapped to that region, it is simply an SPI device, and the "address" is simply an offset from the start of the memory device itself. MCUs with a Quad or Octo-SPI controller are memory mapped. QSPI RAM is not that common and relatively expensive. QSPI is more commonly used for flash memory.
The external memory region can be used by STM32 parts with an FSMC (Flexible
Static Memory Controller) or an FMC (Flexible Memory Controller), or and mentions a QPSI interface. The latter FMC SDRAM, and is generally available on the higher end parts. Apart from the QSPI and NAND flash, these interfaces require using the GPIO EMIF (external memory interface) alternate function to create an address and data bus. So it generally requires parts with high pin count to accommodate. The EMIF can be configured for 8, 16 or 32bit data bus for reduced pin count (and slower access).
Now I can't understand, why you would map the data to this specific
address range, what are the advantages, why don't you just place the
data directly in that huge memory region?
Since it was precipitated by your earlier misconception this question is perhaps redundant, but memory that exists in the memory map can be used to store data accessed as regular variables rather than transferring to an from internal buffers and it can be used as an execution region - code can loaded to and be executed directly from such memory.
Now I'm asking this question because I have a slight feeling I have completely missed the point of what the External RAM region really is.
Self awareness is a skill. That is known as conscious incompetence and is a motivator for learning.
It is my conclusion that the memory regions listed by ARM is showing the full potential of a 32bit MCU, as I incorrectly state that the external RAM region "is quite large!" does not necessarily mean that this is "real" size of that region, if it is even used, it depends on what the vendor can physically achieve within the MCU hardware, and I imagine they would limit hardware capabilities to be competitive on price, power consumption etc.
No, it is largely about the number of pins available for an address bus (except for QSPI). The external memory is a matter for the board design - it is not something the MCU vendor decides must be present. The constraint is a maximum, not a required amount of physical memory. The STM32 FMC supports the following memory sizes/types:
So you can have up to 512Mb of SDRAM for example. The space available for static memories (NOR/PSRAM/SRAM) is significantly larger than the than the typical size of such memories.

Is there any connection between the segments created in memory by a microprocessor and the memory structure of a process in an operating system?

In 8086 microprocessor, we segment the memory into segments of 64K each because of the 16 bit registers (Since a 20 bit address cannot be stored in the 16 bit register). These segments are categorized as code segment, data segment, stack segment and extra segment. This structure is similar to that of created by a process in operating system. Does that mean each process takes up memory equivalent to 4 segments which will be equivalent to 4*64K in case of 8086 ? And if this is true then by doing some more math we can say that only 4 process will be handled by a 8086 microprocessor at a time (i.e. one of the process will be running state and others would be in block or ready state) since maximum of 16 segments are possible (Total memory size / size of each segment = 1MB/64K = 16).
I have just started studying this and saw this equivalence between process and segments. Does any such connection between the segments of the memory and the memory structure of the process actually exists or it's just my crazy imagination ?
A little history helps. Early UNIX(tm) ran on the Digital pdp minicomputer family. The first circulated versions were V6 & V7, which were exclusive to the pdp-11 family. That family could support a whopping 256K of RAM; but the gp register set (used for address formation) were 16bits wide. There was a limited memory protection scheme in the processor, which permitted the kernel (supervisor) to have a separate address space from user (user); and instructions (addresses generated by pc) to be separate from data (generated by other means). This will probably get edited into the dust by pdp-11 fanbois.
At around this time, intel was rolling what was to become the 8086. Current 8-bit CPUs were already straining at a 64K address space limitation, and were using a concept called bank switching to increase that. In bank switching, some sub-ranges of the 64K address space could be re-pointed into a larger memory bank; so although you could carefully address much more memory. The Hitachi 64180 was one of the CPUs that incorporated this into its silicon; most used external memory controllers.
The 8086 addressing scheme was an amalgamation of these notions. You could produce an Operating System which supported dynamically relocated processes and shared text with up to 64K Instruction + 64K Data. The general idea was you take the segment registers out of the programming model, thus if the OS has to relocate the process, it knows that the process had no saved copy of the old segment value. The commercial OS QNX 1.x, 2.x provided this as a model; the later using the 286 extensions to protect against programs that played with the segment registers.
For programs that didn't care about such subtleties (Lotus 123, ...), you could use the segment registers to effectively create a 2^20 address space on the 8086. It is an ugly programming model in this mode because address formation is A=Seg*16+Base, so Seg=1,Base=0 and Seg=0,Base=16 resolve to the same address.
So, you aren't hallucinating, it was quite intentional, if more than a little half-arsed.

How does malloc know where the first available block is in embedded systems?

I have read that malloc has multiple implementations which are platform depended.
How does it work in an embedded device in bare metal programming?
Let's suppose we have an mcu with 256KB FLASH memory and 64KB RAM.
How does it know how much available RAM there is from my program?
For bare metal systems, you'll have a specific segment allocated in the linker script, often called .heap. There is no such thing as memory sharing between processes, meaning that the heap must have a fixed maximum size and therefore is pretty useless in general. malloc doesn't know a thing about how much RAM your program uses since there is no desktop OS in sight.
Your RAM is divided into .stack, .data, .bss and .heap, each with its own fixed maximum size. More about these segments here: https://electronics.stackexchange.com/a/237759/6102. In a typical bare metal MCU application, most of the RAM will be reserved for .data and .bss. You will have something from 128 bytes up to several kb reserved for the stack. You will typically not have a heap at all - but if you do, it will sit there and take up a fixed amount of x kb no matter how much of it you actually use.
malloc in itself could be implemented in different ways indeed. Either you include a "header" together with each allocated segment, the header stating the allocated size and potentially the address of the next available free segment. Or you could implement it as a look-up table where each item is a pointer to the first element and the size.
None of this is particularly relevant, since you shouldn't be using heap allocation in embedded systems. The main reason being that it doesn't make any sense. You don't want arbitrary behavior, you want deteministic behavior. You want to allocate x amount of memory for the worst case and if a heap was to be used it would have to be at least that large anyway, so you gain nothing but bloat from using a heap. Then comes all the usual problems with allocation overhead, fragmentation and leaks.
For bare metal/RTOS applications, do yourself a favour and delete .heap from your linker script, then forget that you ever heard about malloc. A MCU is not a PC.

How to properly assign huge heap space for JVM

Im trying to work around an issue which has been bugging me for a while. In a nutshell: on which basis should one assign a max heap space for resource-hogging application and is there a downside for tit being too large?
I have an application used to visualize huge medical datas, which can eat up to several gigabytes of memory if several imaging volumes are opened size by side. Caching the data to be viewed is essential for fluent workflow. The software is supported with windows workstations and is started with a bootloader, which assigns the heap size and launches the main application. The actual memory needed by main application is directly proportional to the data being viewed and cannot be determined by the bootloader, because it would require reading the data, which would, ultimately, consume too much time.
So, to ensure that the JVM has enough memory during launch we set up xmx as large as we dare based, by current design, on the max physical memory of the workstation. However, is there any downside to this? I've read (from a post from 2008) that it is possible for native processes to hog up excess heap space, which can lead to memory errors during runtime. Should I maybe also sniff for free virtualmemory or paging file size prior to assigning heap space? How would you deal with this situation?
Oh, and this is my first post to these forums. Nice to meet you all and be gentle! :)
Update:
Thanks for all the answers. I'm not sure if I put my words right, but my problem rose from the fact that I have zero knowledge of the hardware this software will be run on but would, nevertheless, like to assign as much heap space for the software as possible.
I came to a solution of assigning a heap of 70% of physical memory IF there is sufficient amount of virtual memory available - less otherwise.
You can have heap sizes of around 28 GB with little impact on performance esp if you have large objects. (lots of small objects can impact GC pause times)
Heap sizes of 100 GB are possible but have down sides, mostly because they can have high pause times. If you use Azul Zing, it can handle much larger heap sizes significantly more gracefully.
The main limitation is the size of your memory. If you heap exceeds that, your application and your computer will run very slower/be unusable.
A standard way around these issues with mapping software (which has to be able to map the whole world for example) is it break your images into tiles. This way you only display the image which is one the screen (or portions which are on the screen) If you need to be able to zoom in and out you might need to store data at two to four levels of scale. Using this approach you can view a map of the whole world on your phone.
Best to not set JVM max memory to greater than 60-70% of workstation memory, in some cases even lower, for two main reasons. First, what the JVM consumes on the physical machine can be 20% or more greater than heap, due to GC mechanics. Second, the representation of a particular data entity in the JVM heap may not be the only physical copy of that entity in the machine's RAM, as the OS has caches and buffers and so forth around the various IO devices from which it grabs these objects.

Can a 32-bit processor load a 64-bit memory address using multiple blocks or registers?

I was doing a little on 32-bit microprocessors and have I have learnt that:
1) A 32-bit microprocessor can only address 2^32 bits of memory which means that the memory pointer size should not exceed 32-bit range i.e. the pointer size should be equal to or less than 32-bit.
2) I also came to know that CPU allocate multiple blocks of memory for things like storing numbers and text, that is up to the program and not related to the size of each address (Source:here).So is it possible that a CPU can use multiple blocks (registers) to store pointers more than 32-bit in size?
Processors can access an essentially unlimited amount of memory by using variations on a technique called bank switching. In a simple bank-switching scheme, the memory chips that are wired to a portion of the address space will have some address inputs fed by the processor and some from an external latching device. Historically, the IBM PC had a 1MB address space, but an expanded memory board would IIRC allow two 16KB regions of that space to be mapped to any of dozens or hundreds of 16KB blocks of memory contained thereon. Nowadays processors generally have a memory-management unit built-in, which maps 4KB or 64KB blocks of memory to any address within a much larger space, and additional circuitry may, with OS support, expand things further.
The big difficulty with bank switching is that any given address might identify many different places in memory depending upon how the bank-switching hardware is configured, so accessing data from memories in a banked region will generally be more complicated than accessing data in directly-accessible memory and will only be possible from code which knows how the bank-switching hardware works. Nowadays it's more common to simply use a processor which can access all the memory one needs, but historically bank-switching was often a useful technique for going beyond processor limitations.
You could store a 64 bit pointer using 2 separate locations in memeory. But it probably wouldn't be useful since your processor can only use 32 bit pointers.