If we know the address space, are we able to compute the machine address-ability? For example, if the address space of a machine is 2^6 bits, what would the machine address-ability be?
In a modern OS, memory space for each process is virtual. The system can manipulate with by far larger amounts of memory than the size of address space. Basically, adressability now only makes sense for a single process, which is limited by size of address space with some reserved regions subtracted.
For example, on Windows 32bit you have rougly 3,5 Gb of address space at process disposal.
We can't determine size of virtual memory available for a process from size of address space because size of reserved regions is OS specific.
The hardware addressable memory and the virtual address space size are two different things. In most processors, the hardware addressable range is significantly smaller than than the virtual addresses size. There are some processors (generally using some kind of segmentation) where the hardware addressable range is larger than the virtual (or programable) address space size.
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Been looking around for answers but none come out satisfying so far, and very often misleading with mixed-up terminologies. Now here's the thing... Physical Memory is the kind of memory;
i. as seen by the CPU as the final memory on its address bus. That means it has already passed the MMU translation / paging, right?
ii. calculated based on the address bus width. If it is 48-bit wide, than it should be 256TB of physical memory addresses available for the system, right?
Now my question;
If I have 8GB of RAM, how on earth the much larger physical memory addresses generated by the CPU are mapped to the physical RAM? What translation unit is there between the Physical memory and the available RAM? From what I read, physical RAM has already passed the virtual / MMU translation phase.
And no, I am not talking about virtual memory. I am asking about the relationships between the physical memory (the one that appears on the CPU address bus) and the actual physical RAM.
Thanks.
As far as I know , the paging system do eliminate external fragment in physical address space, but what about fragment in virtual address space?
In modern OSes the virtual address space is used per process (the kernel has it's own dedicated virtual range), which means that the demands are much lower compared to the whole OS. The virtual address space is usually large enough (2-3 GB per process on x86 and multiple TB (8 on Windows) on x64 machines), so that fragmentation is not such a big issue as for the OS-wide physical address space. Still the issue can arise, especially for long running and memory hungry applications on x86 or other 32 bit architectures. For this the OS provides mechanisms, for example in form of the heap code. An application usually reserves one or more memory ranges as heap(s) when it starts and allocates the required chunks of memory from there later (e.g. malloc). There are a varity of implementations that handle fragmentation of the heap in different ways. Windows provides a special low-fragmentation heap implementation that can be used, if desired. Everything else is usually up to the application or it's libraries.
Let me add a qualification to your statement. Paging systems nearly eliminate fragmentation in the physical address space when the kernel is pageable.
On some systems, the user mode page tables are themselves pageable. On others, they are are physical locations that are not pageable. Then you can get fragmentation.
Fragmentation in the virtual address space tends to occur in heap allocation. The challenge of heap managers is to manage the space while minimizing fragmentation.
I have been reading about how the PCI subsystem gets configured from Bootup, BIOS involvement and mapping of device addresses i.e the BAR's into system Memory.
From the diagram above I am assuming that the address space is physical 4GB RAM with 4GB physical addresses. So, As can be seen above 3GB the device memory is mapped. What happens to this memory on 2GB physical RAM addresses.
If suppose my assumption is wrong and the above map shows virtual address for a 32 bit system. Then how is the device memory mapped to physical addresses for DMA. Is the mapping permanent (non swappable and changeable).
Please help me understand this concept.
If I understand your question, nothing different happens on a 2GB system. There will simply be a "hole" in the physical address space between 2GB and 3GB; that is, there simply won't be a hardware device decoding the addresses here. But otherwise there is no significant difference with respect to PCI devices: they will still be assigned space in the region above 3GB.
It's important to note that the map you show above (physical address space) doesn't necessarily stop at 4GB (since about 1995). Most modern x86 processors have more than 32 address bits. This is why you now often get systems with more than 4GB RAM. And there may be additional holes in the address space above 4GB too.
Actually using the RAM above 4GB requires either the processor's 64-bit mode or PAE (Physical Address Extension) which offers a way to address more than 4GB of physical space in 32-bit mode. [There is also PSE-36 {Page Size Extension} but that's much less commonly used.]
The map you're showing above is specific to physical address space. The x86 virtual address space, (when the processor is operating in 32-bit mode) is 4GB in size, but it does not have all the reserved areas in your diagram. Indeed, the layout of the virtual address space is totally dependent on and determined by the operating system. The usual way that it's configured in linux reserves the part of the virtual address space below the 3GB line for user-mode, and the area above 3GB for kernel-mode. However, this configuration can be changed via the kernel config.
Mapping of the physical address space into virtual address space is managed by the operating system kernel on a page by page basis. A virtual page may be directed either to system RAM or to a PCI device. And note that the page size can vary too, depending on how the processor and page tables are configured.
I thought that virtual address space was a section of RAM allocated to a specific process. But the book I'm reading says that 4 gbs is the standard limit of virtual address space. Isn't that the entire amount of RAM? If that is the case then I'm confused at what virtual address space is. Can anyone enlighten me?
That's the whole point of virtual addresses: The OS handles the physical memory, the process handles its own, virtual memory which is mapped to any memory the OS has available, not necessarily RAM.
On a 32 bit operating system the virtual address space (VAS) is, as you say, usually 4 GiB. 32 bits give you (2^32) addresses (0 ... (2^32)-1), each addressing one byte.
You could have more or less physical RAM and still have a 4-GiB-VAS for each and every process running. If you have less physical RAM, the OS would usually swap to harddrives.
The process doesn't need to know any of this, it can use the full VAS it is given by the OS and it's the OS' job to supply the physical memory.
(This is actually just a dumbed-down version of the Wikipedia article on VAS.)
An excerpt of Wikipedia's article on Physical Address Extension:
x86 processor hardware-architecture is augmented with additional address lines used to select the additional memory, so physical address size increases from 32 bits to 36 bits. This, theoretically, increases maximum physical memory size from 4 GB to 64 GB.
Along with an image explaining the mechanism:
But I can't see how the address space is expanded from 4GB to 64GB. And 4 * 512 * 512 * 4K still equals 4GB, isn't it?
x86 processors running in 32-bit mode uses page translations for memory addresses. This means that there is a mapping layer between the address used by the code (both kernel and user mode) and the actual physical memory. E.g. in Windows all processes map the image of the .exe file to the same address.
The mapping layer between the virtual and physical addresses can normally only map 4GB of memory. With PAE enabled, the 32 bit virtual addresses are mapped 36 bit physical addresses. Still, a single process cannot access more than 4GB at a single time. That's what you see in the image you've pasted, the 32-bit address space of one process. You can also see that the PTE (Page Table Entry) containing the physical address is 64 bit wide.
A PAE aware application can swap in and out different parts of memory into the visible address space to make use of more than 4GB of RAM, but it can only see 4GB at any single point in time.
That's the virtual address space that's still 4GB. The physical address space is larger because the page table entries contain longer physical addresses of pages.
See, the picture says "64-bit PD entry" and "64-bit PT entry". Those extra 32 bits of the entries make up the longer physical addresses of pages.
With this particular scheme your application can still address up to 4GB of memory (minus the kernel portion that's generally inaccessible due to protection) at a time, but if you consider several applications, they can address more than 4GB of memory together.
It does not. The address page never changes. What happens is that via API calls you can SWAP OUT areas of memory against other areas of memory. So, you still have only a full address space of 4gb (2-3 gb usable), but you can have another 2000 blocks of 512mb that you can swap into one part of the address space.