I've looked at a few different articles related to this already but none of them explain the solution in a way that I can understand and replicate. I need to know how to translate a physical address to a virtual address in memory based on the following:
A simple virtual memory system has 32KB physical memory with 16-bit virtual address, of which 12 bits are used as offset. The following is the current content of the page table of one of the processes:
So basically I think the page size of this virtual memory system is 1024KB. I need a process to find the corresponding PA of VA B2A0. If you can give me the process I can go from there, you don't have to give me the final solution :)
Thanks in advance guys. Also, if you know of an article that does this already and I've just missed it, feel free to just link me to that.
Cheers.
32 KB is 2^15.
so there are 15 bits for every physical address, lower 12 of them are used as offset, higher 3 as a number of pageframe.
What virtual page does 0xb2a0 resides in? To determine this, we need to take bits of the address, higher than 2^12. The size of a page is 2^12, that is 4096 or 0x1000, so it is a virtual page number 0xb = 11 (floor of 0xb2a0 / 0x1000). Offset inside the page is 0xb2a0 modulo 0x1000, it's 0x2a0.
Then use the table to translate the virtual page number 11 to a physical pageframe. The virtual page is present (1), and it corresponds to the physical frame number with higher bits 111, that is 111 + twelve 0 in binary, => 0x7000 - it is the address of the start of the physical frame.
Our physical address resides at offset 0x2a0, so, the sought physical address is 0x7000 + 0x2a0 = 0x72a0.
Please, follow this flow and make it clear for you. If you have questions, read the Wikipedia first and if something is still not clear, ask :)
I was trying to do my examination review and study, and I couldn't find a solid answer to this same question. I consolidated what I have learnt, and I hope that whatever I summed up here will help those like me. :)
I find the explanation in the answer above a little hard to understand for my little brain.
I think this link below gives a better overview than Wikipedia's explanation:
http://williams.comp.ncat.edu/addrtrans.htm
This youtube video also offers an excellent guide in explaining the process of virtual address translation:
https://www.youtube.com/watch?v=6neHHkI0Z0o
Back to the question ->>>
The first question is - what is the 'page size' of this virtual memory system?
based on the definition here - https://en.wikipedia.org/wiki/Page_(computer_memory)
I was initially confused between 'pages' and 'page size' but I kinda figured it out now. Pages determines the number of pages available (like in a book), and page size is like (A4,A5,A6 pages in the book!).
As such, since the virtual memory and physical memory offset is the same and is mapped accordingly, we can determine the page size via the offset size. If the offset size is given as 12-bit, then 2^12 = 4,096 Byte a.k.a 4-KB.
Side question for curious minds, how many virtual memory pages are there?
- 16-bit of virtual address space minus 12-bit of offset = 4-bit
- which equals to 2^4 = 16-pages available (thus the table we see!)
Another side question for other curious minds, how many PHYSICAL memory pages are there?
- 32KB of physical Memory = 32 x 1024bytes = 32,768 bytes
- Log(32768) / Log(2) = 15-bits which also means 2^15 for total physical MEMORY
- minus the offset of 12-bit that we already know...
- 15-bit (total physical memory) minus 12-bit (offset) = 3-bit for physical address space
Going to the next question, what is the corresponding physical address of virtual address 0xb2a0 (that is currently set in hex notation)?
#Dmytro Sirenko answer above explains it quite well, I will help to rephrase it here.
We need to remember that our virtual address is - 16-bit, and that address space now contain is value = b2a0 (ignoring the 0x).
My short-cut (please correct me if am wrong), is that since the ratio of the address : offset (page size) is 4:12 = 1:3...
b | 2 a 0
^
page number | offset
When converting hex value b to decimal = 11.
I look into the table, and I found Page Frame = 111 in the table entry number 11.
111 is noted in binary and it correlates to the physical memory frame.
Remember, we were looking at 15-bit of Physical Memory Address space, as such, we can determine that:
1 1 1 | 0 0 0 0 0 0 0 0 0 0 0 0
Address | offset
As Offset are mapped directly from virtual memory to physical memory, we bring the value of (2a0) right into the physical memory. Unfortunately, we can't represent it right away in here because it's in hexadecimal format while my above address space is set in binary.
Considering that I am going to be tested in an examination and I won't be allowed to bring in a calculator... I will do a reverse and answer in Hexadecimal instead. :)
When we convert 111 into decimal (I go by 001 = 1, 010 = 2, 100 = 4, 101 = 5, 110 = 6, 111 = 7).
Now I need to convert from decimal to Hex! = 7 (dec) = 7
As such, the corresponding Physical Memory location of this virtual memory address is.... (loud drums and curtain open....)
7 2 a 0
which is notated in this manner 0x72a0.
Related
"A memory has 1024 storage units with a width of 64. Suppose the memory is byte addressable. What is the address of the highest addressable memory position?"
Please correct me if I'm wrong.
byte addressable means individual bytes in a word have their own addresses.
there are 8 bytes in a 64 bit word.
therefore 8 x 1024 = 8192 addresses overall.
highest address therefore 8191.
I believe this to be true but am not a 100% sure. Please indicate where my logic falters if indeed it does.
I would say 1023.
There are 1024 storage locations, each numbered 0 to 1023, and each storage location holding 64 bits.
So you have a computer where a byte contains 64 bits. A byte is not 8 bits, but the minimum size of a memory location. All modern computers use 8 bits on a byte, but some older computers used 7, 9 og 14 bits on each byte.
But it's a really badly written question, because it does not define what a storage location is. So your interpretation might be right assuming a standard 8 bit in a byte cpu.
What I understand so far is address width is the number of bits in an address.
For example, 4 bits width address can have 2^4 = 16 cases. And what I'm really uncertain is addressability. Based on what I learned is "the size of the most basic unit that can be named by address". So, if we have 4 bits address width and 2 bits addressability, what happens?
I've been really curious about it for a couple of weeks, but still bummer.
Could you guys explain those things by drawing or something?
I think you do get it. There is the number of address bits, the width if you will, and there is the size of the unit those things address. so 8 bits means you have 256 things, 16 bits of address 65536 things. The size of thing is completely independent of the number of address bits. From a programmers perspective almost always we deal in units of bytes, so 8 bits of address would be 256 bytes, 32 bits of address would be 4 gigabytes. As you dig into the logic it is often wasteful to use a byte based address, if you have a peripheral that has 32 bit wide registers and you can only access those as whole 32 bit registers then do you need to connect address line 0 or 1? often not. so at that peripheral the address bus however wide it is (often the whole address bus to the peripheral is a subset of the address bus higher up closer to the processor/software, and those address bits are in units of 32 bit words.
To make things even more confusing, memory parts are often defined in terms of bits, even if they have an 8 or 16 bit data bus. So you might have a 4M part but that is megabits not megabytes...
I am told that I can find the physical address corresponding to a virtual address using /proc/[pid]/pagemap.
I read that this pagemap file is an array of 64-bit entries, with bits 0-54 corresponding to the page frame number. I don't know how to make the leap from this to translating virtual to physical. Partially, I don't know how to find the entry I want in this file; nobody seems to specify how they are indexed.
Also, I don't know if the PFN is virtual or physical. And I don't know what to do with the PFN, regardless. How can I proceed?
Thanks
Divide the VA by the page size (4096 normally), use that as an offset into /proc/self/pagemap. Then take that number (the page), multiply by the pagesize (4096), and offset that by your VA%4094.
Larry
I want to write data to an arbitrary physical memory address to test the error detection and correction feature of my system. One code segment in an existing kernel module is written like this:
u32 addr;
struct page *page;
void *mem;
pci_read_config_dword(priv->mc, I5100_MEMEINJADDRMAT, &addr);
/* Inject error by writing to address */
page = pfn_to_page(addr >> PAGE_SHIFT);
mem = kmap(page) + (addr & (~PAGE_MASK));
*((volatile u32*) (mem)) = 0x01010101;
kunmap(page);
I5100_MEMEINJADDRMAT is the register address of a register in i5100 memory controller. Basically, the memory address is retrieved in that register. I don't understand the remaining code, starting from retrieving a page then perform bitwise operations.
As far as I understand, pfn_to_page is used to get a page that includes a particular physical address by passing in a page frame number as argument. The addr >> PAGE_SHIFT part is to translate from a given address to its corresponding page frame number. But, I don't understand how to use PAGE_SHIFT correctly? What should be the correct data type to use with PAGE_SHIFT?
kmap() returns the appropriate virtual page address then add the offset to get the correct pointer to a virtual memory address. What does (addr & (~PAGE_MASK)) actually do?
My task is to write error injection to a physical address? But the above code seems to write to a virtual address. Is there any other way?
This:
(addr & (~PAGE_MASK))
will clear the bits in addr that are set in PAGE_MASK. Assuming a page size of 4 KB, the PAGE_MASK will likely have its 12 least significant bits set, since 212 = 4096.
So, PAGE_MASK is 0x00000fff. Then, the bit-wise inverse ~PAGE_MASK is simply 0xfffff000, so when addr is bitwise-and:ed with this, the lowest 12 bits of addr are cleared.
/* PAGE_SHIFT determines the page size */
#define PAGE_SHIFT 12
#define PAGE_SIZE (_AC(1,UL) << PAGE_SHIFT)
#define PAGE_MASK (~(PAGE_SIZE-1))
I found these definition in linux-source-3.2.0.
but I get PAGE_MASK is 0xfffff000
so I think this operator is clearing the highest 20bit, or try to get the value of lower 12 bit.
page = pfn_to_page(addr >> PAGE_SHIFT);
It gets the page address using page frame number (page frame number (pfn) is obtained shifting to right the address). Because least significant bits represent offset in the page (12 bits for x86). Page frame number is equal to 20 most significant bits of the physical address for x86.
You can apply shifting on pointer or integer data types. It depends on the situation.
(addr & (~PAGE_MASK)) gets inner offset of the frame. To access correct byte you should add offset to related page.
Virtual address represents physical address. As far as I know, there is no way other than using virtual address. Also, I should say this is kernel virtual space, it is different from userspace virtual addresses.Do not mix userspace with kernel space.
So given I have a 64 bit machine. I know that the memory of the address where my program loads is )
0x8000:0000 and
I need to find the destination memory address such that from 0x8000:0000 , I have at least 512 MB between.
So how do I find out what this destination memory is?
The calculation I did was 8000:0000 + 0x20000000 = result is the destination. But I am not entirely sure if it is correct. So I would need some help in here. Can anyone help me on this?
Something like 0x8000:0000 is typically used to represent a an address in segment:offset notation, when used with a segmented memory architecture. Is that what you're dealing with here? If so, 0x8000:0000 = 0x80000. And 0x80000 + 0x20000000 = 0x20080000.